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Why is SnCl(2) more easly formed than Sn...

Why is `SnCl_(2)` more easly formed than `SnCl_(4)`?

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In the formation of `SaC l_(2)`, the element shows its oxidation state +2 whereas +4 oxidation state is involved when `SnCl_(4)` is to be formed. Due to inert pari effect it isnot so easy for the element to part with its valence s-electron `(5s^(2)sp^(2))` easily whereas valence p- electrons because of lesser penetration can be easily available.
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PbCl_(4) is less stable than SnCl_(4) but PbCl_(2) is more stable than SnCl_(2) . Justify.

Account for the following : (i) Sulphur in vapour from exhibits paramagnetic behavious. (ii) SnCl_(4) is more covalent the SnCl_(2) . (iii) H_(3)PO_(2) is a stronger reducing agent then H_(3)PO_(3) .

Identify the correct statement (s)? A) I-Cl bond is stronger than Br-Br bond B) I-Cl is polar where as Br-Br is non-polar C) \(SnCl_{4}\) more covalent than \(SnCl_{2}\) D) \(O_{2}F_{2}\) oxidises Pu to \(PuF_{6}\)

Which of the following occurs as a consequence of inert pair effect ? a) SnCl_(2) acts as a reducing agent b) SnCl_(4) acts as an oxidising agent c) SnO_(2) is amphoteric d) PbO_(2) is an oxidant e) CCl_(2) is unstable but PbCl_(2) is stable

DINESH PUBLICATION-P BLOCK ELEMENTS (GROUP 13 AND 14 )-Concept based
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