The electronic configurations of four element | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$P_2L, RL, PQ$ and $RQ_2$

Vedclass · Answer

(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:

Similar Questions

Lattice energy of an ionic compound depends upon

The force which holds atoms together in an electrovalent bond is

Write the favourable factors for the formation of ionic bond.

Why does molten sodium chloride conduct electricity?

Which of the following substances is ionic in nature?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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: