Consider the following six changes
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Then calculate value of "", where and are total number of transformations in which magnetic property will be

Question

Consider the following six changes
(i) $NO \longrightarrow NO^+$
(ii) $O_2^- \longrightarrow O_2^{2-}$
(iii) $O_2 \longrightarrow O_2^+$
(iv) $NO^+ \longrightarrow NO^-$
(v) $NO^+ \longrightarrow NO^{2+}$
(vi) $CO \longrightarrow CO^+$
Then calculate value of "$c^3 - b^2 - a$", where $a, b$ and $c$ are total number of transformations in which magnetic property will be changed, bond order increases and bond order decreases respectively.

Prashant Kotian · Accepted Answer

Correct answer: 13

**SOLUTION:**
$A = 5$ (MAGNETIC PROPERTY CHANGES IN ALL EXCEPT $NO^+ \RIGHTARROW NO^{2+}$).
$B = 3$ (BOND ORDER INCREASES IN I, III, VI).
$C = 3$ (BOND ORDER DECREASES IN II, IV, V).
$C^3 - B^2 - A = 3^3 - 3^2 - 5 = 27 - 9 - 5 = 13$.. Step 1: Identify the chemical formulas and geometries: nitrite is NO2^- and nitrile (e.g. organic nitriles R-C≡N or nitryl/nitrile cation NO2^+). Step 2: In NO2^-, Nitrogen has 2 bond pairs and 1 lone pair, leading to a bent (angular) shape. For a nitrile group (R-C≡N) or the nitryl/nitrile cation (NO2^+), the hybridization is sp, giving a linear geometry. Step 3: Consequently, the shapes are angular (bent) and linear respectively, corresponding to option (b).

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