Question

please tel the answer of 3 and 4​

Answer 1

$\huge \underline \bold \red{SOLUTION}:$

1.

$a. \:4x + 6( - 2) = 0 \\ = > x = + 3 \\ \\b. \:1 + 2( + 1) + x + 2( - 2) = 0 \\ = > x = + 1 \\ \\ c.\:x + 3( + 1) = 0 \\ = > x = - 3 \\ \\ d. \:3( + 1) + x + 4( - 2) = 0 \\ = > x = + 5$

therefore in PH3 oxidation state of phosphorous is minimum.

$\underline\bold \red{final \: answer}$

$\boxed {option \: c : PH3}$

_________________________

2.

$2(x) + 7( - 2) = - 4 \\ = > x = + 5$

$\underline\bold \red{final \: answer} \:$

$\boxed{option \: d : + 5}$

hope it helps you