Question

Write about hybridization....in ur own words.Class11 Chemistry...NCERT..​

Answer 1

Explanation:

$\LARGE{\bf{\underline{\underline\color{blue}{GIVEN:-}}}}$

$\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2} = \dfrac{1-cos}{1+cos}$

$\LARGE{\bf{\underline{\underline\color{blue}{SOLUTION:-}}}}$

LHS:

$\sf \to \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}→$

Expand the fractions using (a+b+c)²=a²+b²+c²+2ab+2bc+2ca.

$\sf \to \dfrac{(cos^2-2sincos+sin^2-2cos+2sin+1)}{(cos^2+2sincos+sin^2+2cos+2sin+1)}→$

Rearrange the terms.

$\sf \to \dfrac{(cos^2+sin^2-2sincos-2cos+2sin+1)}{(cos^2+sin^2+2sincos+2cos+2sin+1)}→$

We know that cos²A+sin²A=1.

$\sf \to \dfrac{1-2sincos-2cos}{2sin+1}→$

Now here, take -2cos common from the numerator and +2cos common from the denominator.

$\sf \to \dfrac{1-2cos(sin+2)}{2sin+1}→$

Now, rearrange the terms, add 1 and 1 and take 2 common.

$\to\sf\dfrac{1+1+2sin-2cos}{sin+1}→$

$\to\sf\dfrac{2+2sin-2cos}{sin+1}→$

Take 2 common.

$\to \sf \dfrac{ 2(1+sin) -2cos(sin+1) }{ 2(1+sin) + 2cos(sin +1 ) }→$

$\to \sf{\red{\dfrac{1-cosA}{1+cosA} }}→ </p><p>$

LHS=RHS.

HENCE PROVED!

FUNDAMENTAL TRIGONOMETRIC RATIOS:

$\begin{gathered}\begin{gathered}\boxed{\substack{\displaystyle \sf sin^2 \theta+cos^2 \theta = 1 \\\\ \displaystyle \sf 1+cot^2 \theta=cosec^2 \theta \\\\ \displaystyle \sf 1+tan^2 \theta=sec^2 \theta}}\end{gathered}\end{gathered}$

T-RATIOS:

$\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}$

Answer 2

Answer:

In chemistry, orbital hybridisation (or hybridization) is the concept of mixing atomic orbitals into new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds in valence bond theory.