Class 12th
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New answer posted
7 months agoContributor-Level 10
[CrF6]–4
⇒ Cr+2 → 3d4
F– → WFL → No pairing so unpaired e– = 4
(b) [MnF6]–4
Mn+2 → 3d5
F– → WFL → No pairing unpaired e– = 5
(c) [Cr (CN)6]–4 ⇒ Cr+2 ⇒ d4 CN– → SFL
→ unpaired e– = 2
(d) [Mn (CN)6]–4 ⇒ 3d5,
unpaired e– = 1
New answer posted
7 months agoContributor-Level 10
[V (CO)6]
EAN = 23 – 0 + 2 * 6
= 23 + 12
= 35 ≠ 36
⇒ so it does not obey EAN rule.
New answer posted
7 months agoContributor-Level 10
Acidic nature of hydrides increases down the group in p-block
so H2Te will be most acidic among given options.
New answer posted
7 months agoContributor-Level 10
A jump is seen after 2nd Ip so Ve– = 2
hence configuration would be ns2
New answer posted
7 months agoContributor-Level 10
The quantisation of electric charge, q = ne, applies to electric charge only, even though charge cannot exist without mass.
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