{"id":12543,"date":"2026-05-16T08:45:59","date_gmt":"2026-05-16T08:45:59","guid":{"rendered":"https:\/\/www.vedprep.com\/exams\/?p=12543"},"modified":"2026-05-16T09:47:25","modified_gmt":"2026-05-16T09:47:25","slug":"rotational-spectroscopy-2027","status":"publish","type":"post","link":"https:\/\/www.vedprep.com\/exams\/iit-jam\/rotational-spectroscopy-2027\/","title":{"rendered":"Fundamental concepts of rotational spectroscopy:  IIT JAM 2027"},"content":{"rendered":"<p><span style=\"font-weight: 400;\">Understanding the fundamental concepts of <strong>rotational spectroscopy<\/strong> is crucial for IIT JAM aspirants, as it forms a vital part of physical chemistry. This article will help you grasp the key concepts, including rotational energy, moments of inertia, and molecular symmetry.<\/span><\/p>\n<h2><b>Syllabus and Key Textbooks for Rotational Spectroscopy<\/b><\/h2>\n<p data-path-to-node=\"1\"><strong>Rotational spectroscopy<\/strong> is a core topic in physical chemistry, and it holds a permanent spot right in <b data-path-to-node=\"1\" data-index-in-node=\"102\">Chapter 11<\/b> of the <a href=\"https:\/\/jam2026.iitb.ac.in\/files\/syllabus_CY.pdf\" rel=\"nofollow noopener\" target=\"_blank\"><strong>IIT JAM Physical Chemistry syllabus<\/strong><\/a>. If you want to clear your concepts without drowning in jargon, you can lean on a couple of classic textbooks that we often recommend here at VedPrep:<\/p>\n<ul data-path-to-node=\"2\">\n<li>\n<p data-path-to-node=\"2,0,0\"><b data-path-to-node=\"2,0,0\" data-index-in-node=\"0\">Atkins, P. W. and De Paula, J. <i data-path-to-node=\"2,0,0\" data-index-in-node=\"31\">Physical Chemistry<\/i> (9th edition):<\/b> This one breaks down the core principles and mathematical derivations without making your head spin.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"2,1,0\"><b data-path-to-node=\"2,1,0\" data-index-in-node=\"0\">Levine, I. N. <i data-path-to-node=\"2,1,0\" data-index-in-node=\"14\">Physical Chemistry<\/i> (6th edition):<\/b> If you want a closer look at the rigid rotor model and need to visualize how a molecule actually spins, Levine is your go-to.<\/p>\n<\/li>\n<\/ul>\n<p data-path-to-node=\"3\">Both books give you a solid foundation, which is exactly what you need to tackle those tricky numerical problems in the exam.<\/p>\n<h2><b>Fundamental concepts of rotational spectroscopy For IIT JAM: Rotational Energy and Moments of Inertia<\/b><\/h2>\n<p data-path-to-node=\"6\">When you dive into <strong>rotational spectroscopy<\/strong>, the first thing to get comfortable with is that a molecule\u2019s rotational energy isn&#8217;t continuous\u2014it is <b data-path-to-node=\"6\" data-index-in-node=\"146\">quantized<\/b>. This means a molecule can only spin at specific, approved energy levels.<\/p>\n<p data-path-to-node=\"8\">For a molecule, things are a bit more three-dimensional. We look at three mutually perpendicular axes, giving us three moments of inertia: <span class=\"math-inline\" data-math=\"I_A\" data-index-in-node=\"139\">I<sub>A<\/sub><\/span>, <span class=\"math-inline\" data-math=\"I_B\" data-index-in-node=\"144\">I<sub>B<\/sub><\/span>, and <span class=\"math-inline\" data-math=\"I_C\" data-index-in-node=\"153\">I<sub>C<\/sub><\/span>.<\/p>\n<p data-path-to-node=\"9\">To make things simpler for the exam, we use the <b data-path-to-node=\"9\" data-index-in-node=\"48\">rigid rotor model<\/b>. This model assumes the bonds connecting the atoms are stiff like steel rods and don&#8217;t stretch while spinning. Under this model, you can calculate the energy levels using this formula:<\/p>\n<p data-path-to-node=\"9\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-16663 aligncenter\" src=\"https:\/\/www.vedprep.com\/exams\/wp-content\/uploads\/rigid-rotor-model.png\" alt=\"rigid rotor model\" width=\"222\" height=\"101\" \/><\/p>\n<p data-path-to-node=\"9\">Here, <span class=\"math-inline\" data-math=\"J\" data-index-in-node=\"6\">J<\/span>\u00a0is the rotational quantum number (<span class=\"math-inline\" data-math=\"J = 0, 1, 2...\" data-index-in-node=\"42\">J = 0, 1, 2&#8230;<\/span>), <span class=\"math-inline\" data-math=\"\\hbar\" data-index-in-node=\"59\">h\u00af bar<\/span>\u00a0is the reduced Planck constant (<span class=\"math-inline\" data-math=\"h\/2\\pi\" data-index-in-node=\"97\">h\/2\u03c0<\/span>), and <span class=\"math-inline\" data-math=\"I\" data-index-in-node=\"110\">I<\/span>\u00a0is the moment of inertia. If you can master how <span class=\"math-inline\" data-math=\"E\" data-index-in-node=\"160\">E<\/span> and <span class=\"math-inline\" data-math=\"I\" data-index-in-node=\"166\">I<\/span>\u00a0change relative to each other, you are already halfway to solving most IIT JAM spectral line questions.<\/p>\n<h2><b>Fundamental concepts of rotational spectroscopy For IIT JAM: Classification of Molecular Rotors and Symmetry<\/b><\/h2>\n<p data-path-to-node=\"14\">Not all molecules spin the same way. Based on their shape and symmetry, we group them into four different classes of molecular rotors. This classification tells you exactly how complex a molecule&#8217;s spectrum is going to be.<\/p>\n<ul data-path-to-node=\"15\">\n<li>\n<p data-path-to-node=\"15,0,0\"><b data-path-to-node=\"15,0,0\" data-index-in-node=\"0\">Spherical Top Molecules:<\/b> As per <span style=\"font-weight: 400;\"><strong>rotational spectroscopy, <\/strong>t<\/span>hese are highly symmetrical and have three equal moments of inertia (<span class=\"math-inline\" data-math=\"I_A = I_B = I_C\" data-index-in-node=\"95\">I<sub>A<\/sub> = I<sub>B<\/sub> = I<sub>C<\/sub><\/span>). Think of highly symmetrical shapes like <span class=\"math-inline\" data-math=\"CH_4\" data-index-in-node=\"153\">CH<sub>4<\/sub><\/span>\u00a0or <span class=\"math-inline\" data-math=\"CCl_4\" data-index-in-node=\"161\">CCl<sub>4<\/sub><\/span>. Because they are so perfectly balanced, their rotational spectra are exceptionally neat, showing a series of equally spaced lines.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"15,1,0\"><b data-path-to-node=\"15,1,0\" data-index-in-node=\"0\">Linear Molecules:<\/b> These are straight-line molecules like <span class=\"math-inline\" data-math=\"CO_2\" data-index-in-node=\"57\">CO<sub>2<\/sub><\/span> or <span class=\"math-inline\" data-math=\"HCl\" data-index-in-node=\"65\">HCl<\/span>. For these, the moment of inertia along the molecular axis is essentially zero (<span class=\"math-inline\" data-math=\"I_A = 0\" data-index-in-node=\"149\">I<sub>A<\/sub> = 0<\/span>), while the other two axes are equal (<span class=\"math-inline\" data-math=\"I_B = I_C\" data-index-in-node=\"195\">I<sub>B<\/sub> = I<sub>C<\/sub><\/span>). They have simple rotational spectra and follow very strict selection rules.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"15,2,0\"><b data-path-to-node=\"15,2,0\" data-index-in-node=\"0\">Symmetrical Top Molecules:<\/b> Here, two moments of inertia are equal, but the third one is different (<span class=\"math-inline\" data-math=\"I_A = I_B \\neq I_C\" data-index-in-node=\"99\">I<sub>A<\/sub> = I<sub>B<\/sub> \u2260 I<sub>C<\/sub><\/span>). Molecules like <span class=\"math-inline\" data-math=\"NH_3\" data-index-in-node=\"135\">NH<sub>3<\/sub><\/span> or <span class=\"math-inline\" data-math=\"CH_3Cl\" data-index-in-node=\"143\">CH<sub>3<\/sub>Cl<\/span>\u00a0fit this description. Their spectra are moderately complex but still manageable with a bit of practice.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"15,3,0\"><b data-path-to-node=\"15,3,0\" data-index-in-node=\"0\">Asymmetrical Top Molecules:<\/b> These are the chaotic ones. All three moments of inertia are completely different (<span class=\"math-inline\" data-math=\"I_A \\neq I_B \\neq I_C\" data-index-in-node=\"111\">I<sub>A<\/sub> \u2260 I<sub>B<\/sub> \u2260 I<sub>C<\/sub><\/span>). Water (<span class=\"math-inline\" data-math=\"H_2O\" data-index-in-node=\"142\">H<sub>2<\/sub>O<\/span>) and methanol (<span class=\"math-inline\" data-math=\"CH_3OH\" data-index-in-node=\"162\">CH<sub>3<\/sub>OH<\/span>) belong here. Because they lack symmetry, their rotational spectra look incredibly messy and are tougher to analyze.<\/p>\n<\/li>\n<\/ul>\n<h2><b>Worked Example: Rotational Spectroscopy of a Diatomic Molecule<\/b><\/h2>\n<p data-path-to-node=\"18\">Let\u2019s look at how this plays out in a typical exam question.<\/p>\n<p data-path-to-node=\"19\">Imagine a hypothetical diatomic molecule. Let&#8217;s say its moment of inertia is <span class=\"math-inline\" data-math=\"2.5\\text{ u }\\text{\u00c5}^2\" data-index-in-node=\"77\">2.5 u \u00c5<sup>2<\/sup><\/span> (where <span class=\"math-inline\" data-math=\"\\text{u}\" data-index-in-node=\"108\">u<\/span> is the unified atomic mass unit). The energy levels are calculated using <span class=\"math-inline\" data-math=\"E_J = BJ(J+1)\" data-index-in-node=\"190\">E<sub>J<\/sub> = BJ(J+1)<\/span>, where <span class=\"math-inline\" data-math=\"B\" data-index-in-node=\"211\">B<\/span>\u00a0is the rotational constant. We find <span class=\"math-inline\" data-math=\"B\" data-index-in-node=\"249\">B<\/span>\u00a0using the formula:<\/p>\n<p data-path-to-node=\"19\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-16664 aligncenter\" src=\"https:\/\/www.vedprep.com\/exams\/wp-content\/uploads\/Diatomic-Molecule.png\" alt=\"Diatomic Molecule\" width=\"147\" height=\"97\" \/><\/p>\n<p data-path-to-node=\"19\">The selection rule for a rotational transition is \u0394<span class=\"math-inline\" data-math=\"\\Delta J = \\pm 1\" data-index-in-node=\"50\"> J = \u00b11<\/span>, meaning a molecule can only jump up or down by one level at a time.<\/p>\n<p data-path-to-node=\"22\"><strong>Question<\/strong><\/p>\n<p data-path-to-node=\"23\">A diatomic molecule with a moment of inertia of <span class=\"math-inline\" data-math=\"2.5\\text{ u }\\text{\u00c5}^2\" data-index-in-node=\"48\">2.5 u \u00c5<sup>2<\/sup><\/span> shows a rotational transition line at a frequency of <span class=\"math-inline\" data-math=\"10.5\\text{ GHz}\" data-index-in-node=\"125\">10.5 GHz<\/span>, which corresponds to the <span class=\"math-inline\" data-math=\"J = 1 \\rightarrow J = 2\" data-index-in-node=\"167\">J = 1 \u2192 J = 2<\/span> transition. Calculate the rotational constant <span class=\"math-inline\" data-math=\"B\" data-index-in-node=\"237\">B<\/span>\u00a0and the moment of inertia <span class=\"math-inline\" data-math=\"I\" data-index-in-node=\"265\">I<\/span> in <span class=\"math-inline\" data-math=\"\\text{kg m}^2\" data-index-in-node=\"270\">kg m<sup>2<\/sup><\/span>.<\/p>\n<p data-path-to-node=\"24\"><strong>Solution<\/strong><\/p>\n<p data-path-to-node=\"25\">The energy difference (\u0394<span class=\"math-inline\" data-math=\"\\Delta E\" data-index-in-node=\"23\">E<\/span>) between the <span class=\"math-inline\" data-math=\"J = 1\" data-index-in-node=\"45\">J = 1<\/span> and <span class=\"math-inline\" data-math=\"J = 2\" data-index-in-node=\"55\">J = 2<\/span>\u00a0levels is:<\/p>\n<div data-path-to-node=\"26\">\n<div class=\"math-block\" data-math=\"\\Delta E = E_2 - E_1 = B[2(2+1)] - B[1(1+1)] = 6B - 2B = 4B\">\u0394E = E<sub>2<\/sub> &#8211; E<sub>1<\/sub> = B[2(2+1)] &#8211; B[1(1+1)] = 6B &#8211; 2B = 4B<\/div>\n<div data-math=\"\\Delta E = E_2 - E_1 = B[2(2+1)] - B[1(1+1)] = 6B - 2B = 4B\">\n<p data-path-to-node=\"27\">Wait, let&#8217;s double-check the transition math carefully!<\/p>\n<ul data-path-to-node=\"28\">\n<li>\n<p data-path-to-node=\"28,0,0\">For <span class=\"math-inline\" data-math=\"J=2\" data-index-in-node=\"4\">J=2<\/span>, <span class=\"math-inline\" data-math=\"E_2 = B(2)(3) = 6B\" data-index-in-node=\"9\">E<sub>2<\/sub> = B(2)(3) = 6B<\/span>.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"28,1,0\">For <span class=\"math-inline\" data-math=\"J=1\" data-index-in-node=\"4\">J=1<\/span>, <span class=\"math-inline\" data-math=\"E_1 = B(1)(2) = 2B\" data-index-in-node=\"9\">E<sub>1<\/sub> = B(1)(2) = 2B<\/span>.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"28,2,0\">So, \u0394<span class=\"math-inline\" data-math=\"\\Delta E = 6B - 2B = 4B\" data-index-in-node=\"4\">E = 6B &#8211; 2B = 4B<\/span>.<\/p>\n<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>The frequency (<span class=\"math-inline\" data-math=\"\\nu\" data-index-in-node=\"15\">\u03bd<\/span>) of the transition relates to energy by \u03bd<span class=\"math-inline\" data-math=\"\\nu = \\frac{\\Delta E}{h}\" data-index-in-node=\"59\">\u00a0= \u0394E\/h<\/span>. So, <span class=\"math-inline\" data-math=\"10.5\\text{ GHz} = \\frac{4B}{h}\" data-index-in-node=\"89\">10.5 GHz = 4B\/h<\/span>, which means:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-16665 aligncenter\" src=\"https:\/\/www.vedprep.com\/exams\/wp-content\/uploads\/transition-relates.png\" alt=\"transition relates\" width=\"257\" height=\"140\" \/><\/p>\n<p>Using Planck&#8217;s constant (<span class=\"math-inline\" data-math=\"h = 6.626 \\times 10^{-34}\\text{ J s}\" data-index-in-node=\"25\">h = 6.626 \u00d710<sup>-34 <\/sup>\u00a0J s<\/span>), the frequency value gives us <span class=\"math-inline\" data-math=\"B\" data-index-in-node=\"93\">B<\/span> directly in terms of Hz if we use \u03bd<span class=\"math-inline\" data-math=\"\\nu = 4B\" data-index-in-node=\"129\"> = 4B<\/span>:<\/p>\n<p><img loading=\"lazy\" loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-16666 aligncenter\" src=\"https:\/\/www.vedprep.com\/exams\/wp-content\/uploads\/frequency-value-300x64.png\" alt=\"frequency value\" width=\"300\" height=\"64\" srcset=\"https:\/\/www.vedprep.com\/exams\/wp-content\/uploads\/frequency-value-300x64.png 300w, https:\/\/www.vedprep.com\/exams\/wp-content\/uploads\/frequency-value.png 442w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Now, we can isolate the moment of inertia <span class=\"math-inline\" data-math=\"I\" data-index-in-node=\"42\">I<\/span>:<\/p>\n<p><img loading=\"lazy\" loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-16667 aligncenter\" src=\"https:\/\/www.vedprep.com\/exams\/wp-content\/uploads\/inertia.png\" alt=\"inertia\" width=\"166\" height=\"82\" \/><\/p>\n<p>Plugging in our values:<\/p>\n<p><img loading=\"lazy\" loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-16668 aligncenter\" src=\"https:\/\/www.vedprep.com\/exams\/wp-content\/uploads\/values-300x46.png\" alt=\"values\" width=\"300\" height=\"46\" srcset=\"https:\/\/www.vedprep.com\/exams\/wp-content\/uploads\/values-300x46.png 300w, https:\/\/www.vedprep.com\/exams\/wp-content\/uploads\/values.png 581w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<h2><b>Misconception: Rotational Spectroscopy is only for Spherical Molecules<\/b><\/h2>\n<p data-path-to-node=\"41\">A common trap many students fall into is thinking that rotational spectroscopy only works for perfectly spherical molecules. It is easy to see why people get confused\u2014textbooks love using spherical or simple linear molecules because their calculations don&#8217;t take up three pages of scratch paper.<\/p>\n<p data-path-to-node=\"42\">But here is the reality: <strong>rotational spectroscopy<\/strong> can be applied to almost any molecule with a permanent dipole moment, no matter how lopsided it is.<\/p>\n<p data-path-to-node=\"43\">While asymmetric molecules have wildly complex spectra because of their three unequal moments of inertia, they are still fair game. In fact, physical chemists love analyzing these complex spectra because they act like a unique fingerprint. As per <span style=\"font-weight: 400;\"><strong>rotational spectroscopy, <\/strong>t<\/span>hey give us precise details about rotational constants and centrifugal distortion constants, which tell us exactly how much a bond stretches when a molecule spins too fast<\/p>\n<h2><b>Application: Rotational Spectroscopy in Atmospheric Science<\/b><\/h2>\n<p data-path-to-node=\"46\">To see this theory in action, look at how scientists study the air we breathe. Imagine a fictional research station high up in the Himalayas, trying to track invisible changes in the air currents. The researchers can&#8217;t just stick a regular thermometer into the upper stratosphere to see what&#8217;s happening. Instead, they beam radiation through the air and read the rotational spectra of the gas molecules.<\/p>\n<p data-path-to-node=\"47\">Because every molecule absorbs microwave radiation at specific quantized intervals, the atmospheric scientists can pinpoint exactly which gases are present in the air. By looking at the sharpness and intensity of these spectral lines, they can calculate the exact concentration and temperature of greenhouse gases like <span class=\"math-inline\" data-math=\"CO_2\" data-index-in-node=\"319\">CO<sub>2<\/sub><\/span> and <span class=\"math-inline\" data-math=\"CH_4\" data-index-in-node=\"328\">CH<sub>4<\/sub><\/span>\u00a0without ever leaving the ground. It is also an essential tool for monitoring pollution levels and building accurate climate models.<\/p>\n<h2><b>Exam Strategy: Tips for IIT JAM Aspirants<\/b><\/h2>\n<p data-path-to-node=\"50\">If you want to ace the rotational spectroscopy questions in the IIT JAM, you need a clear strategy. This isn&#8217;t a topic where rote memorization will save you. You need to understand the mechanics of the formulas.<\/p>\n<p data-path-to-node=\"51\">Here is a quick checklist for your revision sessions:<\/p>\n<ul data-path-to-node=\"52\">\n<li>\n<p data-path-to-node=\"52,0,0\"><b data-path-to-node=\"52,0,0\" data-index-in-node=\"0\">Master the Selection Rules:<\/b> Memorize the conditions required for a molecule to show a microwave spectrum (like having a permanent dipole moment) and the transition rule (\u0394<span class=\"math-inline\" data-math=\"\\Delta J = \\pm 1\" data-index-in-node=\"170\"> J = \u00b11<\/span>).<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"52,1,0\"><b data-path-to-node=\"52,1,0\" data-index-in-node=\"0\">Learn the Line Spacing:<\/b> For a rigid diatomic rotor, remember that the separation between consecutive lines in the spectrum is equal to <span class=\"math-inline\" data-math=\"2B\" data-index-in-node=\"135\">2B<\/span>. This single fact solves a massive chunk of numerical problems.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"52,2,0\"><b data-path-to-node=\"52,2,0\" data-index-in-node=\"0\">Watch Your Units:<\/b> Questions often mix up <span class=\"math-inline\" data-math=\"\\text{cm}^{-1}\" data-index-in-node=\"41\">cm<sup>-1<\/sup><\/span>, <span class=\"math-inline\" data-math=\"\\text{Hz}\" data-index-in-node=\"57\">Hz<\/span>, and <span class=\"math-inline\" data-math=\"\\text{Joules}\" data-index-in-node=\"72\">Joules<\/span>. Spend time practicing unit conversions so you don&#8217;t lose easy marks.<\/p>\n<\/li>\n<\/ul>\n<p data-path-to-node=\"53\">At <a href=\"https:\/\/www.vedprep.com\/online-courses\/iit-jam\"><strong>VedPrep<\/strong> <\/a>, we always remind our students that consistent practice beats last-minute cramming. Working through past years&#8217; question papers will show you exactly how the examiners like to frame these problems, helping you build the speed and confidence you need on exam day.<\/p>\n<h2><b>Real-World Example: Rotational Spectroscopy in Molecular Biology<\/b><\/h2>\n<p data-path-to-node=\"56\">While atmospheric scientists use it to look at the sky, molecular biologists sometimes use specialized gas-phase rotational spectroscopy setups to look at the building blocks of life.<\/p>\n<p data-path-to-node=\"58\">By vaporizing the tiny biomolecules into a vacuum gas phase and analyzing their rotational spectra, the team can map out bond lengths, bond angles, and exact molecular symmetry. This gives them a pristine, unaltered look at the molecule&#8217;s structure. In the real world, this helps researchers:<\/p>\n<ul data-path-to-node=\"59\">\n<li>\n<p data-path-to-node=\"59,0,0\">Map out the baseline structures of small amino acids and neurotransmitters.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"59,1,0\">Observe how flexible molecules fold or alter their shape when exposed to subtle environmental shifts.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"59,2,0\">Understand the fundamental structural geometry that drives complex biological interactions.<\/p>\n<\/li>\n<\/ul>\n<p data-path-to-node=\"60\">Even though working with gas-phase biomolecules requires high-resolution instrumentation, the structural insights gained are unmatched.<\/p>\n<h2 data-path-to-node=\"60\"><strong>Conclusion\u00a0<\/strong><\/h2>\n<p data-path-to-node=\"63\">Rotational spectroscopy might seem like an abstract mix of quantum mechanics and math at first, but once you break down the core ideas\u2014quantized energy, moments of inertia, and molecular symmetry\u2014everything starts to click. Mastering the rigid rotor model, line spacings (<span class=\"math-inline\" data-math=\"2B\" data-index-in-node=\"272\">2B<\/span>), and selection rules will give you a massive advantage when tackling Chapter 11 questions on exam day.<\/p>\n<p data-path-to-node=\"63\">To know more in detail from our faculty, watch our YouTube video:<\/p>\n<p class=\"responsive-video-wrap clr\"><iframe title=\"NMR Spectroscopy Organic Chemistry | Electromagnetic Radiation| EMR Chemistry |CSIR NET\/GATE\/IIT JAM\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/0L8ZnvUrmgA?list=PLdZcCa6mtW21HIKg00wd36u8HMjiHnA12\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/p>\n<h2><b>Frequently Asked Questions<\/b><\/h2>\n<p><span style=\"font-weight: 400;\"><br \/>\n<\/span><span style=\"font-weight: 400;\"><style>#sp-ea-16672 .spcollapsing { height: 0; overflow: hidden; transition-property: height;transition-duration: 300ms;}#sp-ea-16672.sp-easy-accordion>.sp-ea-single {margin-bottom: 10px; border: 1px solid #e2e2e2; }#sp-ea-16672.sp-easy-accordion>.sp-ea-single>.ea-header a {color: #444;}#sp-ea-16672.sp-easy-accordion>.sp-ea-single>.sp-collapse>.ea-body {background: #fff; color: #444;}#sp-ea-16672.sp-easy-accordion>.sp-ea-single {background: #eee;}#sp-ea-16672.sp-easy-accordion>.sp-ea-single>.ea-header a .ea-expand-icon { float: left; color: #444;font-size: 16px;}<\/style><div id=\"sp_easy_accordion-1778920664\">\n<div id=\"sp-ea-16672\" class=\"sp-ea-one sp-easy-accordion\" data-ea-active=\"ea-click\" data-ea-mode=\"vertical\" data-preloader=\"\" data-scroll-active-item=\"\" data-offset-to-scroll=\"0\">\n\n<!-- Start accordion card div. -->\n<div class=\"ea-card ea-expand sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-166720\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse166720\" aria-controls=\"collapse166720\" href=\"#\"  aria-expanded=\"true\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-minus\"><\/i> What is the fundamental requirement for a molecule to show a pure rotational spectrum?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse collapsed show\" id=\"collapse166720\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-166720\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>For a molecule to be microwave active (show a pure rotational spectrum), it must possess a permanent dipole moment. When a molecule with a permanent dipole rotates, it generates an oscillating electric field that can interact with the electric field component of microwave radiation. This is why homonuclear diatomic molecules like <span class=\"math-inline\" data-math=\"H_2\" data-index-in-node=\"332\">H<sub>2<\/sub><\/span>, <span class=\"math-inline\" data-math=\"N_2\" data-index-in-node=\"337\">N<sub>2<\/sub><\/span>, and <span class=\"math-inline\" data-math=\"O_2\" data-index-in-node=\"346\">O<sub>2<\/sub><\/span> are rotationally inactive, while heteronuclear ones like <span class=\"math-inline\" data-math=\"HCl\" data-index-in-node=\"407\">HCl<\/span> and <span class=\"math-inline\" data-math=\"CO\" data-index-in-node=\"415\">CO<\/span>\u00a0are active.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-166721\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse166721\" aria-controls=\"collapse166721\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Why is rotational spectroscopy also called microwave spectroscopy?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse166721\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-166721\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>It is simply down to the energy gap. The energy required to transition between different rotational energy levels is relatively small. This small energy gap corresponds perfectly to the photon energies found in the microwave region of the electromagnetic spectrum (frequencies around <span class=\"math-inline\" data-math=\"10^9\" data-index-in-node=\"284\">10<sup>9<\/sup><\/span><sup>\u00a0<\/sup>to <span class=\"math-inline\" data-math=\"10^{11}\\text{ Hz}\" data-index-in-node=\"292\">10<sup>11<\/sup> Hz<\/span>).<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-166722\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse166722\" aria-controls=\"collapse166722\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What exactly is the 'rigid rotor model' and why do we use it?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse166722\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-166722\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>The rigid rotor model is a simplified framework where we assume a rotating molecule behaves like two masses connected by a completely stiff, unbendable rod. In this model, the bond length remains absolutely constant no matter how fast the molecule spins. We use it at VedPrep because it simplifies the quantum mechanical equations and gives a clean, baseline formula (<span class=\"math-inline\" data-math=\"E = BJ(J+1)\" data-index-in-node=\"368\">E = BJ(J+1)<\/span>) that works beautifully for low rotational energy states.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-166723\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse166723\" aria-controls=\"collapse166723\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What is the unit of the rotational constant $B$?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse166723\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-166723\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>In physics-heavy problems, <span class=\"math-inline\" data-math=\"B\" data-index-in-node=\"27\">B<\/span>\u00a0is often calculated in Joules (<span class=\"math-inline\" data-math=\"\\text{J}\" data-index-in-node=\"60\">J<\/span>) or Hertz (<span class=\"math-inline\" data-math=\"\\text{Hz}\" data-index-in-node=\"80\">Hz<\/span>). However, in standard IIT JAM physical chemistry problems, it is incredibly common to see <span class=\"math-inline\" data-math=\"B\" data-index-in-node=\"181\">$B$<\/span> expressed in wavenumbers, specifically <span class=\"math-inline\" data-math=\"\\text{cm}^{-1}\" data-index-in-node=\"222\">cm<sup>-1<\/sup><\/span>. Always pay close attention to whether you need to divide your formula by the speed of light (<span class=\"math-inline\" data-math=\"c\" data-index-in-node=\"331\">c<\/span>) to match the unit demands of the question.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-166724\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse166724\" aria-controls=\"collapse166724\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What exactly is the 'rigid rotor model' and why do we use it?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse166724\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-166724\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>The rigid rotor model is a simplified framework where we assume a rotating molecule behaves like two masses connected by a completely stiff, unbendable rod. In this model, the bond length remains absolutely constant no matter how fast the molecule spins. We use it at VedPrep because it simplifies the quantum mechanical descriptions and gives a clean, baseline behavior that works beautifully for low rotational energy states before real-world complications set in.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-166725\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse166725\" aria-controls=\"collapse166725\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What is the selection rule for rotational transitions?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse166725\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-166725\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>For a simple rotor, the gross selection rule is that the molecule must possess a permanent dipole moment. The specific quantum mechanical selection rule dictates that the molecule can only transition to the immediately adjacent higher energy level during absorption, or the immediately adjacent lower level during emission. It cannot skip levels during a pure rotational transition.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-166726\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse166726\" aria-controls=\"collapse166726\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> How does isotopic substitution affect the rotational spectrum of a molecule?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse166726\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-166726\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>When you replace an atom in a molecule with its heavier isotope (like swapping carbon-12 for carbon-13), the chemical properties and bond length remain identical, but the total mass increases. A heavier mass increases the molecule's resistance to spinning (its moment of inertia). Because the rotational constant is inversely related to this resistance, the energy levels compress, and the entire rotational spectrum shifts to lower frequencies.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-166727\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse166727\" aria-controls=\"collapse166727\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What happens to a molecule's rotation when it reaches very high rotational states?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse166727\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-166727\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>In reality, molecules aren't perfectly rigid. When a molecule spins incredibly fast at high rotational states, centrifugal force pulls the atoms outward. This increases the bond length, which changes the molecule's resistance to spinning. To correct for this real-world stretching, scientists introduce a correction factor known as the centrifugal distortion constant, which accounts for the slight lowering of energy levels at high speeds.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-166728\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse166728\" aria-controls=\"collapse166728\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Why does the intensity of spectral lines first rise and then fall as the rotational quantum states increase?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse166728\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-166728\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>The intensity of a spectral line depends directly on how many molecules occupy the initial energy state. Two opposing factors dictate this population: the degeneracy of the state (the number of available spatial orientations, which increases for higher states) and the thermal distribution factor (which decreases exponentially as the energy of the state increases). The interplay of these two factors creates a curve where the population\u2014and thus line intensity\u2014peaks at a specific intermediate state before fading out.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-166729\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse166729\" aria-controls=\"collapse166729\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Are spherical top molecules like methane active in pure rotational spectroscopy?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse166729\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-166729\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>No, they are not. Even though methane can rotate dynamically on multiple axes, its perfect tetrahedral symmetry means its net dipole moment is exactly zero. Because it lacks a permanent dipole moment to interact with the radiation, it cannot give a pure rotational spectrum.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-1667210\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse1667210\" aria-controls=\"collapse1667210\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What physical information do we actually gain by analyzing a rotational spectrum?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse1667210\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-1667210\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>The primary prize of rotational spectroscopy is calculating the exact molecular resistance to spinning from the observed spacing of the spectral lines. Once you have that value and know the masses of the individual atoms involved, you can calculate the precise bond lengths and bond angles of the molecule with incredible accuracy.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-1667211\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse1667211\" aria-controls=\"collapse1667211\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Why are rotational spectral lines often broadened in high-pressure gas samples?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse1667211\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-1667211\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>When gas pressure is high, molecules crowd together and collide frequently. These rapid collisions shorten the lifespan of a molecule in any particular rotational state. According to quantum mechanical uncertainty principles, a shorter lifetime in a state creates a larger uncertainty in the energy measurement, which visually smears and broadens the spectral lines. This is why researchers prefer low-pressure gas samples.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-1667212\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse1667212\" aria-controls=\"collapse1667212\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Can we study the rotational transitions of liquids or solids?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse1667212\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-1667212\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>Practically speaking, no. In liquids and solids, molecules are packed so tightly that intermolecular forces and constant collisions completely disrupt free molecular rotation. Instead of clean, sharp individual lines, the spectrum turns into a smeared, uninterpretable blur. True rotational spectroscopy requires gas-phase samples where molecules have the space to spin freely.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-1667213\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse1667213\" aria-controls=\"collapse1667213\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> How do I distinguish between symmetric top and asymmetric top molecules for the exam?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse1667213\" data-parent=\"#sp-ea-16672\" role=\"region\" aria-labelledby=\"ea-header-1667213\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>Look at the components of their structural symmetry along the three primary axes. If the molecule is symmetrical enough that its rotational resistance is identical along two axes but different along the third (like a pyramid shape), it is a symmetric top. If the molecule is completely asymmetrical such that its rotational resistance is entirely different along all three axes (like a bent water molecule), it's an asymmetric top.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<\/div>\n<\/div>\n<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Understanding the fundamental concepts of rotational spectroscopy is crucial for IIT JAM aspirants, as it forms a vital part of physical chemistry. This article will help you grasp the key concepts, including rotational energy, moments of inertia, and molecular symmetry.<\/p>\n","protected":false},"author":11,"featured_media":12542,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":"","rank_math_seo_score":89},"categories":[23],"tags":[7408,7410,7411,861,7412,7409,2922],"class_list":["post-12543","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-iit-jam","tag-fundamental-concepts-of-rotational-spectroscopy-for-iit-jam","tag-fundamental-concepts-of-rotational-spectroscopy-for-iit-jam-notes","tag-fundamental-concepts-of-rotational-spectroscopy-for-iit-jam-questions","tag-physical-chemistry","tag-rotational-spectroscopy-for-iit-jam","tag-spectroscopy","tag-vedprep","entry","has-media"],"acf":[],"_links":{"self":[{"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/posts\/12543","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/users\/11"}],"replies":[{"embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/comments?post=12543"}],"version-history":[{"count":6,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/posts\/12543\/revisions"}],"predecessor-version":[{"id":16674,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/posts\/12543\/revisions\/16674"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/media\/12542"}],"wp:attachment":[{"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/media?parent=12543"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/categories?post=12543"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/tags?post=12543"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}