Correct $$statement(s)$$ among (I) to (IV) regarding group $$14$$ elements is (are)I) Silicon exists in diamond form and does not have an allotrope which resembles graphite.$$II)$$ All elements other than carbon in this group can expand their octet.$$III)$$ Hybridization of Sn in [$$Sn(OH)6$$ ]$$2$$− is $$sp3d2$$.$$IV)$$ Lead does not from any tetrahalides while all other elements of the group form.

Question

Correct $$statement(s)$$ among (I) to (IV) regarding group $$14$$ elements is (are)I) Silicon exists in diamond form and does not have an allotrope which resembles graphite.$$II)$$ All elements other than carbon in this group can expand their octet.$$III)$$ Hybridization of Sn in [$$Sn(OH)6$$ ]$$2$$− is $$sp3d2$$.$$IV)$$ Lead does not from any tetrahalides while all other elements of the group form.

Infinity Learn · Accepted Answer

$$I)$$ Correct. As silicon cannot for pi bond involving $$p-orbitals$$, it does not have a graphite analogue.$$II)$$ Correct. They all have empty $$d-orbitals$$ to expand their octet.$$III)$$ Correct.$$IV)$$ Lead forms tetrahalides except iodide.

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