Chemical Bonding - Result Question 1
1. The isoelectronic set of ions is
(2019 Main, 10 April I)
(a) $F^{-}, Li^{+}, Na^{+}$ and $Mg^{2+}$
(b) $N^{3-}, Li^{+}, Mg^{2+}$ and $O^{2-}$
(c) $Li^{+}, Na^{+}, O^{2-}$ and $F^{-}$
(d) $N^{3-}, O^{2-}, F^{-}$ and $Na^{+}$
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Answer:
Correct Answer: 1. (d)
Solution:
- Key Idea: Isoelectronic species contains same number of electrons.
The species with its atomic number and number of electrons are as follows :
| Species (ions) | At. no. $(\boldsymbol{Z})$ | No. of electrons |
|---|---|---|
| $N^{3-}$ | 7 | $7+3=10$ |
| $O^{2-}$ | 8 | $8+2=10$ |
| $F^{-}$ | 9 | $9+1=10$ |
| $Na^{+}$ | 11 | $11-1=10$ |
| $Li^{+}$ | 3 | $3-1=2$ |
| $Mg^{2+}$ | 12 | $12-2=10$ |
Thus, option (d) contains isoelectronic set of ions.
BETA
