The electronic configurations of four element | vedclass

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(C) First,determine the valency of each element based on its electronic configuration:$L (1s^2, 2s^2 2p^4)$ has $6$ valence electrons,so it gains $2$

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$P_2L, RL, PQ$ and $RQ_2$

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(C) First,determine the valency of each element based on its electronic configuration:$L (1s^2, 2s^2 2p^4)$ has $6$ valence electrons,so it gains $2$ electrons to complete its octet. Valency = $-2$.$Q (1s^2, 2s^2 2p^6, 3s^2 3p^5)$ has $7$ valence electrons,so it gains $1$ electron. Valency = $-1$.$P (1s^2, 2s^2 2p^6, 3s^1)$ has $1$ valence electron,so it loses $1$ electron. Valency = $+1$.$R (1s^2, 2s^2 2p^6, 3s^2)$ has $2$ valence electrons,so it loses $2$ electrons. Valency = $+2$.Combining these to form neutral ionic compounds:$P (+1)$ and $L (-2)$ form $P_2L$.$R (+2)$ and $L (-2)$ form $RL$.$P (+1)$ and $Q (-1)$ form $PQ$.$R (+2)$ and $Q (-1)$ form $RQ_2$.Thus,the correct set of formulas is $P_2L, RL, PQ$ and $RQ_2$.

The electronic configurations of four elements $L$,$P$,$Q$,and $R$ are given as: $L (1s^2, 2s^2 2p^4)$; $Q (1s^2, 2s^2 2p^6, 3s^2 3p^5)$; $P (1s^2, 2s^2 2p^6, 3s^1)$; $R (1s^2, 2s^2 2p^6, 3s^2)$. The formulas of the ionic compounds that can be formed between these elements are:

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