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NEET Chemistry Solved Question Paper 2006

Multiple Choice Questions

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

$\frac{hm}{4 π ∆ ν}$
$\frac{4 π}{hm ∆ ν}$
$\frac{h}{4 πm ∆ v}$
$\frac{4 πm}{h ∆ ν}$

$\frac{h}{4 πm ∆ v}$

According to Heisenberg's uncertainty principle

$∆$ x ^. $∆$ p $⩾$ $\frac{h}{4 π}$ ( $∆$ p = m $∆ ν$ )

$∆$ x ^. m $∆$ v $⩾$ $\frac{h}{4 π}$

$∆$ x = $\frac{h}{4 πm ∆ v}$

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

57.3 kJ

$\underset{(strong acid)}{HCl}$ + $\underset{(strong base)}{KOH}$ $\overset{neutralisation}{⇌}$ 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 .

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

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 .

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

Mg(OH)₂ $⇌$ Mg²⁺ + 2OH^-

the solubility product K_sp 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

Which of the following transitions have minimum wavelengths ?

n₄ $\to$ n₁
n₂ $\to$ n₁
n₄ $\to$ n₂
n₃ $\to$ n₁

n₄ $\to$ n₁

E = $\frac{hc}{λ}$ or E $\propto$ $\frac{1}{λ}$

Thus a decrease in wave length represents an increase in energy For n₄ $\to$ n₁transition .These are greater the. energy difference and lesser will be wavelength .Thus , for n₄ $\to$ n₁ transition has minimum wavelength .

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

12 h

We know ,

N_t = N_o x ${(\frac{1}{2})}^{n}$

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

N_o = initial amount

n = number of half lives elapsed

$\frac{N_{t}}{N_{o}}$ = ${(\frac{1}{2})}^{n}$

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

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

n = 6

Time taken (T) = t_1/2 x n = 2 x 6 = 12 h

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

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 .

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

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

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

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 .