NEET chemistry

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Question
CBSEENCH11008366

Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is 

  • Mg< Ca < Cl < P

  • Cl < P < Mg < Ca

  • P < Cl < Ca < Mg

  • Ca < Mg < P < Cl

Solution

B.

Cl < P < Mg < Ca

With the increase in the number of electron in the same shell, the atomic radii decrease due to increase in effective nuclear charge. However, atomic radii increases, as the number of shells increases. Thus, on moving down a group atomic radii increases.
The electronic configuration of the given elements is
Mg12 = [Ne] 3s2
Ca20 = [Ar] 4s2
P15 = [Ne]3s2 3p3
Cl17 = [Ne] 3s2 3p5
In Mg, P and Cl, the number of electrons is increasing in the same shell thus, the order of their atomic radii is
Cl <P < Mg
In Ca, the electron is entering in the higher shell, thus, it has the highest atomic radii among the given. Thus, the order of radii is
Cl < P < Mg < Ca

Sponsor Area

Question
CBSEENCH11008367

The reaction,

2A (g) + B (g)  ⇌ 3C (g) + D (g) 

is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of  D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression

  • [(0.75)3 (0.25)] / [(1.00)2 (1.00)]

  • [(0.75)3 (0.25) ] / [(0.50)2(0.75)]

  • [(0.75)3(0.25)] / [(0.50)2(0.75)]

  • [(0.75)3(0.25)] - [(0.75)2(0.25)]

Solution

B.

[(0.75)3 (0.25) ] / [(0.50)2(0.75)]

space space space space space space space space space space space space space space space space 2 space straight A space left parenthesis straight g right parenthesis space space plus space straight B space left parenthesis straight g right parenthesis space space space space rightwards harpoon over leftwards harpoon space space space space space space space 3 straight C space left parenthesis straight g right parenthesis space plus space straight D space left parenthesis straight g right parenthesis
initial space space space space space space space space space space space space 1 space space space space space space space space space space space space space space space 1 space space space space space space space space space space space space space space space space space space space space space space 0 space space space space space space space space space space space space space 0
At space equil space space space space 1 minus 0.50 space space space space space space space space 1 minus 0.25 space space space space space space space space space space 75 space space space space space space space space space space space space 0.25

straight K space equals space fraction numerator left parenthesis 0.75 right parenthesis cubed left parenthesis 0.25 right parenthesis over denominator left parenthesis 0.50 right parenthesis squared left parenthesis 0.75 right parenthesis end fraction

Question
CBSEENCH11008368

The pressure exerted by 6.0 g of methane gas in a 0.03 m3 vessel at 129o C is (Atomic masses C = 12.01, H = 1.01 and R = 8.314 JK-1 mol-1)

  • 215216 Pa

  • 13409 Pa

  • 41648 Pa

  • 31684 Pa

Solution

C.

41648 Pa

Given,
Volume , V = 0.03 m3
Temperature, T = 129 + 273 = 402 K
mass of methane, w = 6.0 g
mol. Mass of methane, M = 12.01 + 4 x 1.01 = 16.05
From, ideal gas equation,
pV = nRT
straight p space equals fraction numerator 6 over denominator 16.05 end fraction space straight x space fraction numerator 8.314 space straight x space 402 over denominator 0.03 end fraction space space equals space 41648 space Pa

Question
CBSEENCH11008369

Among the following which one has the highest cation to anion size ratio?

  • CsI

  • CsF

  • LiF

  • NaF

Solution

B.

CsF

The order of size of given cations is 
Li+ < Na+ < Cs+
and the order of the size of given anions is 


I- > F-
Thus, when the cation is largest and anion is smallest the cation to anion size ratio is maximum. Hence cation to anion size ratio is maximum

Question
CBSEENCH11008370

Three moles of an ideal gas expanded spontaneously into the vacuum. The work done will be

  • infinte

  • 3 J

  • 9 J

  • zero

Solution

D.

zero

W = pext. ΔV
For vaccum,
ptext = 0
therefore,
W = 0  ΔV = 0