﻿ For an electron , if the uncertainty in velocity is ∆v , the uncertainty in its position (∆x)  is given by : from Chemistry NEET Year 2006 Free Solved Previous Year Papers

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# NEET Chemistry Solved Question Paper 2006

#### Multiple Choice Questions

1.

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 a gas , at constant temperature the volume of a gas is inversely proportional to its pressure" .

$\propto$ $\frac{1}{\mathrm{P}}$

or PV = constant

on differentiating the equation

d(PV) = d        (constant)

= PdV + VdP =0

= VdP = - PdV

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

2.

Which of the following sequence is correct as per Aufbau 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 are filled with electrons in the increasing order of their energies , "i.e. , sub-shell of lower energy will 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

3.

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 region of definite shape around the nucleus where the probability of finding the electron is maximum .Therefore , an atom has a both characteristic energy and a characteristic shape .

4.

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 converts into a anion .

Thus the formation ofionic bond is favoured by

(I) Low ionisation potential of metal

(II) Greater valtie of electron affinity of non metal

(III) Higher value of lattice energy of the resulting ionic compound .

5.

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

• 3

• 9

• 5

• 8

B.

9

Mg(OH)2 $⇌$ Mg2+ + 2OH-

the solubility product Ksp of

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

6.

Which of the following transitions have minimum 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 an increase in energy For n4 $\to$ ntransition .These are greater the. energy difference and lesser will be wavelength .Thus , for n4 $\to$ n transition has minimum wavelength .

7.

The enthalpy change ($∆$H) for the neutralisation of M HCl by caustic potash in dilute solution at 298 K is :

• 68 kJ

• 65 kJ

• 57.3 kJ

• 50 kJ

C.

57.3 kJ

+  $\stackrel{\mathrm{neutralisation}}{⇌}$ KCl + H2O

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

# 8.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

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

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

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

9.

A radioactive sample is emitting 64 times radiations than non-hazardous limit .If its half-life is 2 hours , after what time it becomes non- hazardous ?

• 16 h

• 10 h

• 12 h

• 8 h

C.

12 h

We know ,

Nt = No x ${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

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

10.

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 than maximum value of energy depending upon freauency of incident radiations .