The electronic configurations of four element | vedclass

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

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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 valence shell configuration:$L$ has $6$ valence electrons $(2s^2 2p^4)$,so it gains $2$ electrons to complete its octet,valency = $-2$.$Q$ has $7$ valence electrons $(3s^2 3p^5)$,so it gains $1$ electron to complete its octet,valency = $-1$.$P$ has $1$ valence electron $(3s^1)$,so it loses $1$ electron,valency = $+1$.$R$ has $2$ valence electrons $(3s^2)$,so it loses $2$ electrons,valency = $+2$.Now,combine them to form neutral ionic compounds:Between $P(+1)$ and $L(-2)$: $P_2L$Between $R(+2)$ and $L(-2)$: $RL$Between $P(+1)$ and $Q(-1)$: $PQ$Between $R(+2)$ and $Q(-1)$: $RQ_2$Thus,the correct formulae are $P_2L, RL, PQ$ and $RQ_2$.

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

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The electronic configurations of four elements $L$,$P$,$Q$,and $R$ are given in the brackets:$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 formulae of ionic compounds that can be formed between these elements are: