Class 11th

4.21. According to VSEPR theory, if CH₄ were square planar, the bond angle would be 90°. For tetrahedral structure, the bond angle is 109°28? Therefore, in square planar structure, repulsion between bond pairs would be more and thus the stability will be less.

Moreover, the orbitals of carbon in ground and excited states look as shown below. In the excited state, the electrons from 4 H atoms occupy one 2s orbital and three 2p orbitals. This results in an sp³ hybridisation, which is the hybridisation of a tetrahedral geometry

4.19. N₂ < SO₂ < ClF₃ < K₂O

The ionic character in a molecule is dependent upon the electronegativity difference between the constituting atoms. The greater the difference, the greater will be the ionic character of the molecule.

4.17. Electronegativity: Electronegativity is the tendency of an atom to attract shared pair of electrons towards itself. Electronegativity of any given element is never the same.

It depends on the element it is bonded with in a compound. Electronegativity is a relative quantity and cannot be measured.
Whereas, electron gain enthalpy is the change in enthalpy when an electron is added to a neutral gaseous atom to form an anionic specie.

It can have a positive or negative value. Every element has a specific electron gain enthalpy value.

4.16. Significance of dipole moment are:

In predicting the nature of the molecules: Molecules with specific dipole moments are polar in nature and those of zero dipole moments are non-polar in nature.

In the determination of shapes of molecules.

In calculating the percentage ionic character.

4.15. In CO₂, there are two C=O bonds. Each C=O bond is a polar bond. The net dipole moment of CO₂ molecule is zero. This is possible only if CO₂ is a linear molecule. (O=C=O). The bond dipoles of two C=O bonds cancel the moment of each other.

Whereas, H₂O has a net dipole moment of 1.84 D. H₂O molecule has a bent structure because here the O—H bonds are oriented at an angle of 104.5° and do not cancel the bond moments of each other.

 

velocity of a freely falling body is v= $\sqrt[]{2gh}$

And $?=\frac{h}{mv}=\frac{h}{m\sqrt[]{2gh}}$

$?=h$ -1

 The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
a) 1.2 nm (b) 1.2 x 10-3 nm
(c) 1.2 x 10-6  nm (d). 1.2 x 10 nm

(d)

We can write

OS + PS > OP …….(i)

OS > PS – OP …….(ii)

$\left|\stackrel{?}{a}-\stackrel{?}{b}\right|$ $>$ $\left|\stackrel{?}{a}\right|$ + $\left|\stackrel{?}{b}\right|$ ….(iii)

The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus of both sides as:

$\left|\left|\stackrel{?}{a}-\stackrel{?}{b}\right|\right|$ > $\left|\left|\stackrel{?}{a}\right|-\mathrm{}\mathrm{}\left|\stackrel{?}{b}\right|\mathrm{}\right|$ ….(iv)

If the two vectors $\stackrel{?}{a}$ and $\stackrel{?}{b}$ act along a straight line but in the opposite direction, then we can write $\left|\stackrel{?}{a}-\stackrel{?}{b}\right|$ = $\left|\stackrel{?}{a}\right|$ - $\left|\stackrel{?}{b}\right|\dots ..\left(v\right)$

Combining (iv) and (v), we get

$\left|\stackrel{?}{a}-\stackrel{?}{b}\right|$ $\ge$ $\left|\stackrel{?}{a}\right|$ - $\left|\stackrel{?}{b}\right|$