A) \[NH_{4}^{+}\] and \[NH_{2}^{-}\]
B) \[CH_{3}^{-}\] and \[CH_{3}^{+}\]
C) \[SO_{4}^{2-}\],\[PO_{4}^{3-}\] and \[{{[B{{F}_{4}}]}^{-}}\]
D) \[NH_{4}^{+}\] and \[N{{H}_{3}}\]
Correct Answer: C
Solution :
[c] Hybridisation can be calculated by calculating the no of valence electron and dividing it by 8. In \[S{{O}_{4}}^{2-}=\text{Total}\,\,\text{no}\text{.}\,\,\text{of}\,{{e}^{-}}\] \[=6+(6\times 4)+2=32\] So, no. of hybrid orbitals \[=\frac{32}{8}=4\] \[\therefore \,\,\,s{{p}^{3}}\] hybridization. Similarly, for \[P{{O}_{4}}^{3-};\] no. of hybrid orbitals \[=\frac{5+24+3}{8}=\frac{32}{8}=4\] Hybridisation \[=s{{p}^{3}}\] Similarly, for \[BF_{4}^{-}\], it is \[s{{p}^{3}}\].
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