A) \[s{{p}^{2}},\text{ }s{{p}^{3}}\] and \[s{{p}^{2}}\] respectively
B) \[sp,\text{ }s{{p}^{2}}\] and \[s{{p}^{3}}\] respectively
C) \[s{{p}^{2}},\text{ }sp\] and \[s{{p}^{3}}\] respectively
D) \[s{{p}^{2}},\text{ }s{{p}^{3}}\] and sp respectively
Correct Answer: B
Solution :
| [b] \[NO_{2}^{+}=\frac{1}{2}[5+0+0-1]=2sp;\] |
| \[NO_{2}^{-}=\frac{1}{2}[5+0+1-0]=3sp;\] |
| \[NH_{4}^{+}=\frac{1}{2}[5+4+0-1]=4s{{p}^{3}}\] |
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