{"id":9323,"date":"2026-06-22T10:23:00","date_gmt":"2026-06-22T04:53:00","guid":{"rendered":"https:\/\/1000sciencefairprojects.com\/tech\/inorganic-chemistry-exceptions-quiz-jee-main\/"},"modified":"2026-06-22T13:04:39","modified_gmt":"2026-06-22T07:34:39","slug":"inorganic-chemistry-exceptions-quiz-jee-main","status":"publish","type":"post","link":"https:\/\/1000sciencefairprojects.com\/tech\/inorganic-chemistry-exceptions-quiz-jee-main\/","title":{"rendered":"Inorganic chemistry exceptions quiz for JEE Main revision"},"content":{"rendered":"
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Updated 18 June 2026 \u00b7 7 min read \u00b7 India<\/div>\n
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How to use this quiz<\/p>\n

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This is a quick revision quiz on the exceptions and anomalies in inorganic chemistry<\/a><\/strong> that show up again and again in JEE Main. Read each question, decide your answer, then tap the card to reveal the answer and a short reason. Aim to explain the why in one line, that is what separates a safe score from a top score.<\/p>\n<\/div>\n<\/div>\n

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On this page<\/p>\n

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  1. Anomalous electron configurations<\/a><\/li>\n
  2. Ionisation enthalpy exceptions<\/a><\/li>\n
  3. Electron gain enthalpy and inert pair<\/a><\/li>\n
  4. Anomalies of period 2 elements<\/a><\/li>\n
  5. Mixed exceptions<\/a><\/li>\n
  6. Score yourself<\/a><\/li>\n<\/ol>\n<\/div>\n

    Anomalous electron configurations<\/h2>\n
    \nQ1. Why is the configuration of chromium [Ar] 3d5 4s1 and not 3d4 4s2 Tap to reveal<\/span><\/summary>\n
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    Answer:<\/strong> Because a half filled 3d5 set with 4s1 is more stable than 3d4 4s2.<\/p>\n

    Why:<\/strong> Exactly half filled and fully filled subshells gain extra stability from symmetric charge distribution and greater exchange energy, so one 4s electron shifts into 3d.<\/p>\n<\/div>\n<\/details>\n

    \nQ2. Why is copper [Ar] 3d10 4s1 rather than 3d9 4s2 Tap to reveal<\/span><\/summary>\n
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    Answer:<\/strong> Because a completely filled 3d10 with 4s1 is more stable than 3d9 4s2.<\/p>\n

    Why:<\/strong> Fully filled d subshells maximise exchange energy and symmetry, so copper adopts 3d10 4s1. The same logic explains the silver and gold anomalies.<\/p>\n<\/div>\n<\/details>\n

    Ionisation enthalpy exceptions<\/h2>\n
    \nQ3. Why is the first ionisation enthalpy of boron lower than that of beryllium Tap to reveal<\/span><\/summary>\n
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    Answer:<\/strong> Because boron loses a 2p electron while beryllium loses a more tightly held 2s electron.<\/p>\n

    Why:<\/strong> The 2p electron in boron is higher in energy and better shielded by the filled 2s, so it is easier to remove, making boron lower despite a higher nuclear charge.<\/p>\n<\/div>\n<\/details>\n

    \nQ4. Why is the first ionisation enthalpy of oxygen lower than that of nitrogen Tap to reveal<\/span><\/summary>\n
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    Answer:<\/strong> Because nitrogen has a stable half filled 2p3 configuration.<\/p>\n

    Why:<\/strong> Removing an electron from nitrogen breaks a stable half filled set, which is hard. Oxygen loses a paired 2p electron and is relieved of pairing repulsion, so its value is lower.<\/p>\n<\/div>\n<\/details>\n

    Electron gain enthalpy and inert pair<\/h2>\n
    \nQ5. Why is the electron gain enthalpy of chlorine more negative than that of fluorine Tap to reveal<\/span><\/summary>\n
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    Answer:<\/strong> Because the small, compact 2p subshell of fluorine causes strong electron electron repulsion.<\/p>\n

    Why:<\/strong> Fluorine is tiny, so the incoming electron faces high repulsion in the dense 2p shell. Chlorine is larger, the added electron is less crowded, so chlorine releases more energy.<\/p>\n<\/div>\n<\/details>\n

    \nQ6. What is the inert pair effect and one example Tap to reveal<\/span><\/summary>\n
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    Answer:<\/strong> It is the reluctance of the ns2 pair to take part in bonding down a group, for example Tl prefers +1 and Pb prefers +2.<\/p>\n

    Why:<\/strong> Down group 13 to 15 the ns2 electrons are held more tightly due to poor shielding by d and f electrons, so lower oxidation states (Tl+1, Pb+2, Bi+3) become more stable.<\/p>\n<\/div>\n<\/details>\n

    Anomalies of period 2 elements<\/h2>\n
    \nQ7. Why does nitrogen not form pentahalides such as NCl5 while phosphorus forms PCl5 Tap to reveal<\/span><\/summary>\n
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    Answer:<\/strong> Because nitrogen has no available d orbitals, so its maximum covalency is four.<\/p>\n

    Why:<\/strong> Phosphorus can expand its octet using 3d orbitals to reach covalency five. Nitrogen, limited to the second shell, cannot, so NCl5 does not exist.<\/p>\n<\/div>\n<\/details>\n

    \nQ8. Why does fluorine show only the minus one oxidation state and no positive states Tap to reveal<\/span><\/summary>\n
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    Answer:<\/strong> Because it is the most electronegative element and has no d orbitals to expand its octet.<\/p>\n

    Why:<\/strong> With the highest electronegativity and no accessible d orbitals, fluorine cannot be oxidised by other elements, so it is restricted to minus one.<\/p>\n<\/div>\n<\/details>\n

    \nQ9. Give one example of a diagonal relationship and the reason behind it Tap to reveal<\/span><\/summary>\n
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    Answer:<\/strong> Lithium resembles magnesium because of similar charge to size ratio and polarising power.<\/p>\n

    Why:<\/strong> On moving across and down, size and polarising power can stay close, so pairs like Li and Mg, Be and Al, B and Si share properties. For example, lithium and magnesium both form nitrides and have covalent character in their halides.<\/p>\n<\/div>\n<\/details>\n

    Mixed exceptions<\/h2>\n
    \nQ10. Why is the atomic radius of gallium slightly smaller than that of aluminium Tap to reveal<\/span><\/summary>\n
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    Answer:<\/strong> Because of poor shielding by the intervening 3d electrons in gallium.<\/p>\n

    Why:<\/strong> The 3d10 electrons shield the nuclear charge poorly, so the effective nuclear charge on gallium is higher, pulling its outer electrons in and offsetting the expected increase.<\/p>\n<\/div>\n<\/details>\n

    \nQ11. Why do zirconium and hafnium have almost identical atomic radii Tap to reveal<\/span><\/summary>\n
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    Answer:<\/strong> Because of the lanthanoid contraction.<\/p>\n

    Why:<\/strong> The steady decrease in size across the lanthanoids cancels the normal increase expected from hafnium being a period lower, so Zr and Hf end up nearly the same size and behave alike.<\/p>\n<\/div>\n<\/details>\n

    \nQ12. Why is the bond enthalpy order F2 less than Cl2 even though F is above Cl Tap to reveal<\/span><\/summary>\n
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    Answer:<\/strong> Because of strong lone pair repulsion in the small fluorine molecule.<\/p>\n

    Why:<\/strong> The two fluorine atoms are so small that their non bonding lone pairs repel strongly, weakening the F to F bond, so the F2 bond enthalpy is lower than that of Cl2, an important anomaly.<\/p>\n<\/div>\n<\/details>\n

    Score yourself<\/h2>\n

    Count one mark for the correct answer and one more if you could state the reason in a sentence.<\/p>\n

    \n\n\n\n\n\n\n\n
    Score (out of 24)<\/th>\nWhat it means<\/th>\n<\/tr>\n<\/thead>\n
    20 to 24<\/td>\nStrong. You can defend the why, not just the what.<\/td>\n<\/tr>\n
    14 to 19<\/td>\nGood base. Revisit ionisation and inert pair logic.<\/td>\n<\/tr>\n
    Below 14<\/td>\nRe-read these anomalies and try again in two days.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n
    Revision tip.<\/strong> Exceptions are high yield because examiners love them. Keep a one page sheet of anomalies and read it the night before, then the morning of the exam.<\/div>\n

    Looking for another way to challenge your brain beyond chemistry. If you are a student interested in technology and problem solving, take a look at our guide<\/a> to the FCRF Hackathon 2026<\/a>, a cyber and digital forensics challenge for students and professionals. For more study resources, visit the 1000 Science Fair Projects home page<\/a>.<\/p>\n

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    Keep reading<\/span><\/p>\n