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JEE Main Mock Test Series 2020 & Previous Year Papers

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Question 1: The compound(s) formed upon combustion of sodium metal in excess air is (are)    [IIT JEE ]
a) Na2O2
b) Na2O
c) NaO2
d) NaOH
Answer: a & b.
Solution: Na burns in oxygen to form sodium oxide (Na2O) which further absorbs oxygen to form sodium peroxide (Na2O2). So, Na2O and Na2Oare formed when Na metals burns in excess of oxygen.

Question 2:
STATEMENT-1: Alkali metals dissolves in liquid ammonia to give blue solution because
STATEMENT-2: Alkali metals in liquid ammonia give solvated species of the type [M(NH3)n]+ (M = alkali metals).    [IIT JEE ]
a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
c) Statement-1 is True, Statement-2 is False
d) Statement-1 is False, Statement-2 is True
Answer: b
Solution: Alkali metals dissolves in liquid ammonia to give solvated complex ion  [M(NH3)n] and  solvated electrons. The solvated electrons provide deep blue colour to the solution.
Hence, the correct option is b.

Question 3: The set of representing the correct order of first ionization potential is    [IIT JEE]
a) K > Na > Li
b) Be > Mg > Ca
c) B > C > N
d) Ge > Si > C
Answer: b
Solution: The first ionization energy of  elements of same group decreases down the group due to increase in the size to atom because of which the effective nuclear charge on the outermost electron decreases and it requires less energy to remove it from the shell. Be , Mg  &  Ca are all alkali earth metals and are the member of 2nd group of modern periodic table. Order of atomic radii for these elements is Be < Mg < Ca. So the order of first ionization potential would be Be > Mg > Ca.
Hence, the correct option is b.

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