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Write Brief Answer · Q21

Q.Explain the following, give appropriate reasons.

(i) Ionisation potential of N is greater than that of O
(ii) First ionisation potential of C-atom is greater than that of B atom, where as the reverse is true is for second ionisation potential.
(iii) The electron affinity values of Be and Mg are almost zero and those of N (0.02 eV) and P (0.80 eV) are very low
(iv) The formation of F⁻
(g) from F(g) is exothermic while that of O²⁻(g) from O
(g) is endothermic.
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Step 1 (i). Nitrogen's configuration 2p32p^3 is exactly half-filled, an extra-stable arrangement; oxygen's 2p42p^4 has one pair of electrons in the same orbital, adding extra electron-electron repulsion that makes its electron comparatively easier to remove. So IE(N)>IE(O)IE(N) > IE(O), even though O has one more proton.

Step 2 (ii). For IE1, carbon (2s22p22s^2 2p^2) simply has a higher nuclear charge than boron (2s22p12s^2 2p^1) with no anomaly favouring boron, so IE1(C)>IE1(B)IE_1(C) > IE_1(B) follows the normal trend. For IE2: removing a second electron takes B+B^+ (2s22s^2, filled and stable) to B2+B^{2+} -- breaking a stable filled subshell -- while it takes C+C^+ (2s22p12s^2 2p^1, an ordinary, not-yet-stable configuration much like boron's own ground state) to C2+C^{2+}, an easier removal. So for the SECOND ionisation the comparison reverses: IE2(B)>IE2(C)IE_2(B) > IE_2(C).

Step 3 (iii). Be (2s22s^2) and Mg (3s23s^2/ns2ns^2) already have a completely filled outer s-subshell; adding an electron would have to enter a new, higher-energy subshell, which is unfavourable, giving near-zero electron affinity. N (2p32p^3) and P (3p33p^3) have an exactly half-filled p-subshell; adding an electron would pair up an electron in an already-stable, symmetric arrangement, which is also unfavourable (though slightly less so for P, whose larger 3p orbital has less electron-electron repulsion than N's compact 2p), giving very low (but slightly higher than Be/Mg) electron affinity. …

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