Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

NEET Chemistry Solved Question Paper 2006

Multiple Choice Questions

Which of the following transitions haveminimum 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 anincrease in energy Forn₄ $\to$ n₁transition .These are greater the. energy difference andlesser will be wavelength .Thus , forn₄ $\to$ n₁ transition has minimum wavelength .

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

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

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 = $\frac{h}{4 πm ∆ v}$

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 thanmaximum value of energy depending uponfreauency of incident radiations .

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

57.3 kJ

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

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

Mg(OH)₂ $⇌$ Mg²⁺ + 2OH^-

the solubility product K_spof

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

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 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 .

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

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 .