Question

As $\%s$ - character of a hybrid orbital increases

• A. Bond angle decreases
• B. Bond strength decreases
• C. Bond length decreases
• D. Electronegativity of orbital decreases

Answer 1

The correct option is C

Bond length decreases

Let's think about the character of a hybrid orbital.

We know, in overlapping

$s-s$ > $s-p$ > $p-p$

Now, as s character increases in hybrid orbital, it will have bigger front lobes and more tendancy to overlap.

$\%S$ character and bond angle

${sp}^3$ $\Rightarrow$ s - character $\to 25\%$

$\Rightarrow$Tetrahedral $\Rightarrow$ Bond angle $\to {109.5}^{\circ }$

[Because 4 orbitals will be formed]

${sp}^2$$\Rightarrow$ s - character $\to 33.33\%$

$\Rightarrow$Trigonal planar $\Rightarrow$Bond angle $\to$ ${120}^{\circ }$

[Because 3 orbitals will be formed]

$sp\Rightarrow$ s - character $\to 50\%$

$\Rightarrow$Linear $\Rightarrow$ Bond angle $\Rightarrow {180}^{\circ }$

[Because 2 orbitals will be formed]

$\Rightarrow As\%$ s character increases, bond angle increases.

$\%s$ character $\&$ bond strength

As we have discussed, as s - character increases, it will have bigger front lobes and it will have more overlapping.

And we know that more overlapping means more bond strength.

Therefore, as s - character increases, bond strength increases.

$\%s$ character and bond length

Higher bond strength implies smaller bond length because if the bond length is small, it will be hard to break the bond.

$\%s$ character and electronegativity

Electronegativity is the tendency of an atom to attract electrons towards itself in a molecule of a compound.

The electronegativity increases with the increase in 's' character in the hybrid orbital because the 's' orbitals being more near to the nucleus have greater tendency to attract the shared pair of electron.