The oxidation number and the electronic confi | vedclass

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

Question

(D) To find the oxidation number of sulphur $(S)$ in $H_2SO_4$:Let the oxidation number of $S$ be $x$.$2 	imes (+1) + x + 4 	imes (-2) = 0$$2 + x -

Vedclass · Accepted Answer

$+6$; $1s^2 2s^2 2p^6$

Vedclass · Answer

(D) To find the oxidation number of sulphur $(S)$ in $H_2SO_4$:Let the oxidation number of $S$ be $x$.$2 	imes (+1) + x + 4 	imes (-2) = 0$$2 + x - 8 = 0$$x = +6$In $H_2SO_4$,sulphur is in its $+6$ oxidation state,meaning it has lost all its valence electrons $(3s^2 3p^4)$.The electronic configuration of neutral sulphur $(S)$ is $1s^2 2s^2 2p^6 3s^2 3p^4$.After losing $6$ electrons to reach the $+6$ state,the configuration becomes $1s^2 2s^2 2p^6$.

The oxidation number and the electronic configuration of sulphur in $H_2SO_4$ is

Similar Questions

What are the oxidation states of the two sulfur atoms in thiosulfuric acid $(H_2S_2O_3)$?

$M$ is the molecular weight of $KMnO_4$. The equivalent weight of $KMnO_4$ when it is converted into $K_2MnO_4$ is

Oxidation state of oxygen in $F_2O$ is

The oxidation state of nitrogen is incorrectly given for:
Compound $\to$ Oxidation state

Vedclass Test Series

Exam Paper Generator

Online Exam Module