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# 1.Orbital is :circular path around the nucleus in which the electrons revolves space around the nucleus where the probability of finding the electron is maximum amplitude of electron wave none of the above

B.

space around the nucleus where the probability of finding the electron is maximum

An atomic orbital is a three dimensional regionof definite shape around the nucleus where theprobability of finding the electron is maximum .Therefore , an atom has a both characteristicenergy and a characteristic shape .

2.

Which of the following transitions haveminimum wavelengths ?

• n4 $\to$ n1

• n2 $\to$ n1

• n4 $\to$ n2

• n3 $\to$ n1

A.

n4 $\to$ n1

E =$\frac{\mathrm{hc}}{\mathrm{\lambda }}$or E$\propto$$\frac{1}{\mathrm{\lambda }}$

Thus a decrease in wave length represents anincrease in energy Forn4$\to$n1transition .These are greater the. energy difference andlesser will be wavelength .Thus , forn4$\to$n1 transition has minimum wavelength .

3.

A metal surface is exposed to solar radiations :

• the emitted electrons have energy less than a maximum value of energy depending upon frequency of incident radiations

• the emitted electrons have energy less than maximum value of energy depending upon intensity of incident radiations

• the emitted electrons have zero energy

• the emitted electrons have energy equal to energy of photons of incident light

A.

the emitted electrons have energy less than a maximum value of energy depending upon frequency of incident radiations

A metal surface is exposed to solar radiations ,the emitted electrons have energy less thanmaximum value of energy depending uponfreauency of incident radiations .

4.

A radioactive sample is emitting 64 timesradiations than non-hazardous limit .If itshalf-life is 2 hours , after what time it becomesnon- hazardous ?

• 16 h

• 10 h

• 12 h

• 8 h

C.

12 h

We know ,

Nt = Nox${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

where Nt = amount left after expiry of 'n' half lives

No = initial amount

n = number of half lives elapsed

$\frac{{\mathrm{N}}_{\mathrm{t}}}{{\mathrm{N}}_{\mathrm{o}}}$=${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

$\frac{1}{64}$=${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

${\left(\frac{1}{2}\right)}^{6}$=${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

n = 6

Time taken (T) = t1/2 x n = 2 x 6 = 12 h

5.

The enthalpy change ($∆$H) for the neutralisationof M HCl by caustic potash in dilute solution at298 K is :

• 68 kJ

• 65 kJ

• 57.3 kJ

• 50 kJ

C.

57.3 kJ

$\underset{\left(\mathrm{strong}\mathrm{acid}\right)}{\mathrm{HCl}}$+$\underset{\left(\mathrm{strong}\mathrm{base}\right)}{\mathrm{KOH}}$$\stackrel{\mathrm{neutralisation}}{⇌}$KCl + H2O

In this reaction , HCl is the strong acid and KOHis the base and it has been found that the heat ofneutralisation of a strong acid with a strong baseis always constant i.e. , 57.3 kJ .

6.

The Ksp of Mg(OH)2 is 1 x 10-12 , 0.01 M Mg(OH)2will precipitate at the limited pH :

• 3

• 9

• 5

• 8

B.

9

Mg(OH)2$⇌$Mg2+ + 2OH-

the solubility product Kspof

Mg(OH)2 = [Mg2+][OH-]2

1 x 10-12 = 0.01 [OH-]2

[OH-]2 = 1 x 10-10

[OH-] = 10-5

We know

[H+][OH-] = 10­-14

[H+][10-5]= 10-14

[H] = 10-14/10-5 = 10-9

pH = - log[H+] = - log 10-9 = 9

7.

For an electron , if the uncertainty in velocity is$∆$v ,the uncertainty in its position ($∆$x) is given by :

• $\frac{\mathrm{hm}}{4\mathrm{\pi }∆\mathrm{\nu }}$

• $\frac{4\mathrm{\pi }}{\mathrm{hm}∆\mathrm{\nu }}$

• $\frac{\mathrm{h}}{4\mathrm{\pi m}∆\mathrm{v}}$

• $\frac{4\mathrm{\pi m}}{\mathrm{h}∆\mathrm{\nu }}$

C.

$\frac{\mathrm{h}}{4\mathrm{\pi m}∆\mathrm{v}}$

According to Heisenberg's uncertainty principle

$∆$x.$∆$p$⩾$$\frac{\mathrm{h}}{4\mathrm{\pi }}$ ($∆$p = m$∆\mathrm{\nu }$)

$∆$x. m$∆$v$⩾$$\frac{\mathrm{h}}{4\mathrm{\pi }}$

$∆$x =$\frac{\mathrm{h}}{4\mathrm{\pi m}∆\mathrm{v}}$

8.

Ionic compounds are formed most easily with :

• low electron affinity , high ionisation energy

• high electron affinity , low ionisation energy

• low electron affinity , low ionisation energy

• high electron affinity , high ionisation energy

B.

high electron affinity , low ionisation energy

An ionic bond is formed between an electropositive metal and a electronegative non metal .The electropositive metal converts into a cation ,while the electronegative non metal convertsinto a anion .

Thus the formation ofionic bond is favoured by

(I)Low ionisation potential of metal

(II)Greater valtie of electron affinity of nonmetal

(III)Higher value of lattice energy of theresulting ionic compound .

9.

Which of the following sequence is correct as perAufbau principle ?

• 3s < 3d < 4s < 4p

• ls < 2p < 4s < 3d

• 2s < 5s < 4p < 5d

• 2s < 5s < 4p < 5f

B.

ls < 2p < 4s < 3d

According to aufbau principle , "sub-shells arefilled with electrons in the increasing order oftheir energies , "i.e. , sub-shell of lower energywill be filled first with electrons .

Thus correct order is -

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p< 5s < 4d < 5p < 6s < 4f < 5d < 6p< 7s < 5f < 6d ... etc

10.

Equation of Boyle's law is :

• $\frac{\mathrm{dP}}{\mathrm{P}}$ = -$\frac{\mathrm{dV}}{\mathrm{V}}$

• $\frac{\mathrm{dP}}{\mathrm{P}}$ = +$\frac{\mathrm{dV}}{\mathrm{V}}$

• $\frac{{\mathrm{d}}^{2}\mathrm{P}}{\mathrm{P}}$ = -$\frac{\mathrm{dV}}{\mathrm{dT}}$

• $\frac{{\mathrm{d}}^{2}\mathrm{P}}{\mathrm{P}}$ = +$\frac{{\mathrm{d}}^{2}\mathrm{V}}{\mathrm{dT}}$

A.

$\frac{\mathrm{dP}}{\mathrm{P}}$ = -$\frac{\mathrm{dV}}{\mathrm{V}}$

According to Boyle's law , "for a given mass of agas , at constant temperature the volume of a gasis inversely proportional to its pressure" .

V$\propto$$\frac{1}{\mathrm{P}}$

or PV = constant

on differentiating the equation

d(PV) = d (constant)

= PdV + VdP =0

= VdP = - PdV

=$\frac{\mathrm{dP}}{\mathrm{P}}$= -$\frac{\mathrm{dV}}{\mathrm{V}}$