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The hybridisation of atomic orbitals of nitrogen in `NO_2^+ , NO_3^- ` and `NH_4^(+)` are :A. `sp^2, sp^3 and sp^2` respectivelyB. `sp, sp^2 and sp^3` respectivelyC. `sp^2, sp and sp^3` respectivelyD. `sp^2, sp^3 and sp` respectively |
Answer» Correct Answer - B In `NO_2^+` total no. of valence electrons = 5 + 2 x 6-1 =16 Now 16+8 = `2(Q_1) + 0(R_1)` = 2 Thus, type of hybridization is sp. In `NO_3^(-)` total no. of valence electrons =5+ 3 x 6+1=24 Now, 24 + 8 = 3`(Q_1) + 0(R_1)` = 3 Thus type of hybridisation =`sp^2` In `NH_4^(+)` ions, total no of valence electrons = 5+1 x 4-1 =8 Now 8+2 = 4`(Q_1)+0(R_1)=4` Thus type of hybridization is `sp^3` Combining all the results, option (b) is correct. |
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