Question

Topic : VSEPR Theory...

H2O molecule is not linear although it is of AB2 type ( molecules of the type)..

explain through its structure..

Answer 1

A water molecule is not linear because of the electron structure of the oxygen atoms in the water molecules. Oxygen has 6 valence electrons in the second energy level.
hope it will help you dear

Answer 2

Hello!

• L.P. - $\bf{Lone \:Pair}$
• B.P. - $\bf{Bonding \:Pair}$

$\boxed{\boxed{\sf VSEPR\: Theory}}$

$\bf{V}$alence $\bf{S}$hell $\bf{E}$lectron $\bf{P}$air $\bf{R}$epulsion Theory states that as electrons are negatively charged, the $\underline{\sf valence \:electron}$ in different atoms in a molecule $\underline{\sf repel \:each\: other}$.

But,

The $\underline{\sf lone\: pair}$ electrons takes $\underline{\sf more\: space}$ than bonding electrons, as they are only attracted to one atom.

That's why they $\underline{\sf repel \:more\: than\: bonding \:electron}$.

Order of repulsions in different types of e⁻ pairs :-

$\boxed{\sf L.P. - L.P. \:\textgreater \:B.P. - L.P. \: \textgreater \:B.P. - B.P.}$

$\bf{For \:H_{2}O}$ :-

One can't put $\bf{H}$ in the centre because it can only hold two electrons, as its $\underline{\sf Principle \:Quantum}$ number is $\bf{1}$

Therefore,
$\bf{O}$ goes in the centre.

Forming single bonds to each hydrogen leaves two more pairs of e⁻ .

These are lone pairs.
There are $\underline{\sf four \:pairs}$ of e⁻ around the O-atom.

Therefore, $\bf{It\: can \:not\:be\: Linear}$.

If each pair of electrons repelled equally it would be in a tetrahedral arrangement, with $\bf{109^{\circ}}$ bond angles.

But the $\underline{\sf lone\: pairs\: repel\: more}$ than bonding pairs, which compresses the bonding angle to $\bf{104.5^{\circ}}$

Cheers!