JEE Advanced is a high level exam that checks your concept by putting various types of questions. The questions are in a different format from JEE mains.m, herein the questions are : single correct multiple choice questions, multiple correct multiple choice questions, statement based questions and comprehension type questions.
Q1. The size of the second and third row transition elements being almost the same. This is due to
Solution.
(c)
Q2. Out of F_2,Cl_2,Br_2 and I_2, most oxidizing agent isOut of F_2,Cl_2,Br_2 and I_2, most oxidizing agent is
Solution.
(a)
(a)
Q3. Select the correct order of 〖(IE)〗_1
Solution.
(d)
Q4. Of the four ∆H values needed to calculate the lattice energy using the Born-Haber cycle, the one that is most difficult to measure is
Solution
(b)
Q5. The lanthanides contraction refers to
Solution
(b)
Q6. The statement that is not correct for the periodic classification of elements is
Solution.
(d)
Q7.N_0/2 atoms of X(g) are converted into X^+ (g) by energy E_1. N_0/2 atoms of X(g) are converted into X^- (g) by energy E_2. Hence, ionization potential and electron affinity of X(g) are
Solution
(a)
(a)
Q8.Following the transition elements, 〖(IE)〗_1 drops abruptly in Ga, In and Tl. This is due to
Solution
(b)
(b)
Q9.Electronegativity and electron affinity of an element A are X and Y respectively. Hence, ionization potential of A is
Solution
(a) AB_2 has bcc structure, A^(2+) possess face centred cubic lattice. B^- ions occupy all the (100%) tetrahedral voids. Thus each A^(2+) is in contact with B^- and each B^- with 4A^(2+) ions.
(a) AB_2 has bcc structure, A^(2+) possess face centred cubic lattice. B^- ions occupy all the (100%) tetrahedral voids. Thus each A^(2+) is in contact with B^- and each B^- with 4A^(2+) ions.
Q10.Following triads have approximately equal size
Solution
(b) All factual statements
(b) All factual statements