79-Q79- Identify the rows- which are correctly matched - nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- Column I nbsp- nbsp-Column II- nbsp-I- Diamond nbsp-II- Graphite-III- Buckminster fullerene-IV- Carbon dioxide- ndash-sp2 hybridized C- ndash-sp2 hybridized C- ndash-sp2 hybridized C- ndash-sp hybridized C- nbsp- -1- 1 amp- IV only nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- -2- II amp- IV only- nbsp- -3- II- III amp- IV only nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-4- I- II- III amp- IV

Dear Student, Hybridization of allotropes of nbsp;Carbon is nbsp; given below Diamond nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp ...

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Byju's Answer

Standard XII

Chemistry

Ores and Minerals

79 Q79. I...

Question

79

Q79. Identify the rows, which are correctly matched ?

Column I Column II

I. Diamond
II. Graphite
III. Buckminster fullerene
IV. Carbon dioxide –sp² hybridized C
–sp² hybridized C
–sp² hybridized C
–sp hybridized C

(1) 1 & IV only (2) II & IV only

(3) II, III & IV only (4) I, II, III & IV

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Solution

Dear Student,

Hybridization of allotropes of Carbon is given below
Diamond Sp³
Graphite Sp²
Buckminster fullerene Sp²
CO₂ Sp

Regards

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Similar questions

Q. The compounds contaning sp hybridized carbon atom are:

Q. Consider the following anions:

When attached to

s_{p^{3}}

-hybridized carbon, their leaving group ability in nucleophilic substitution reaction decreases in the order:

Column 1: Compound Column

Column 2: Oxidation state Column

Column 3: Type of linkage

$\begin{matrix} Column 1 & Column 2 & Column 3 (I) & N a_{2} S_{2} O_{3} & (i) & + 5 & (P) & S - O - S (I I) & N a_{2} S_{4} O_{6} & (i i) & - 2 & (Q) & - S - S - (I I I) & H_{2} S_{2} O_{7} & (i i i) & + 6 & (R) & - O - O - (I) & H_{2} S_{2} O_{8} & (i v) & + 4 & (S) & S = S \end{matrix}$

The only correct combination in which the hybridization of one of the S atom is $s p^{2}$ :

Q. The correct order of increasing percentage s character in the hybrid orbitals in the given molecules/ions is: (assume all hybrid orbitals are equivalent)

\begin{matrix} C O_{3}^{2 -} & X e F_{4} & I_{3}^{-} & N C l_{3} & B e C l_{2} (I) & (I I) & (I I I) & (I V) & (V) \end{matrix}

Column-I-Compound, Column-II-Oxidation state, Column-III-Type of linkage

$\begin{matrix} C o l u m n 1 & C o l u m n 2 & C o l u m n 3 (I) & N a_{2} S_{2} O_{3} & (i) & + 5 & (P) & S - O - S (I I) & N a_{2} S_{4} O_{6} & (i i) & - 2 & (Q) & S - S (I I I) & H_{2} S_{2} O_{7} & (i i i) & + 6 & (R) & - (O - O) - (I V) & H_{2} S_{2} O_{8} & (i v) & + 4 & (S) & S = S \end{matrix}$

The only correct combination in which the hybridization of one of the S atom is $S p^{2}$ .

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Column I	Column II
I. Diamond II. Graphite III. Buckminster fullerene IV. Carbon dioxide	–sp² hybridized C –sp² hybridized C –sp² hybridized C –sp hybridized C

Dear Student, Hybridization of allotropes of Carbon is given below Diamond Sp3 Graphite Sp2 Buckminster fullerene Sp2 ​CO2 Sp Regards

Dear Student,

Hybridization of allotropes of Carbon is given below
Diamond Sp³
Graphite Sp²
Buckminster fullerene Sp²
CO₂ Sp

Regards