KCET · Chemistry · p Block Elements (Group 15, 16, 17 & 18)

Which is true regarding nitrogen ?

A Less electronegative
B Has low ionisation enthalpy
C d-orbitals are available
D Ability to form \( \mathrm{p} \pi-\mathrm{p} \pi \) bonds with itself

Verified Solution

Answer & Solution

Correct Answer

(D) Ability to form \( \mathrm{p} \pi-\mathrm{p} \pi \) bonds with itself

Step-by-step Solution

Detailed explanation

Electronic configuration for \( \mathrm{N}(Z=7) \) is \( 1 s^{2} 2 s^{2} 2 p^{3} \), so, it does not have d-orbitals. Due to completely half-filled p-
orbital, its ionization enthalpy is high, and nitrogen is third most electronegative element in periodic table. Nitrogen can
form pII - pII bonds to itself.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Match List-I with List-II

List-I (Complex)		List-II (Geometry)
(a)	\([\text{Co(NH}_3\text{)}_6]^{3+}\)	(i)	Trigonal bipyramidal
(b)	\([\text{NiCl}_4]^{2-}\)	(ii)	Octahedral
(c)	\([\text{Ni(CN)}_4]^{2-}\)	(iii)	Tetrahedral
(d)	\([\text{Fe(CO)}_5]\)	(iv)	Square planar

Choose the correct answer from the options given below.

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