Question:medium

Match List I with List II:

Show Hint

Always memorize crystal systems by constraints on (a, b, c) and angles (\(\alpha, \beta, \gamma\)); this makes matching questions direct and fast.
Updated On: Jul 18, 2026
  • A – IV; B – I; C – II; D – III
  • A – IV; B – III; C – II; D – I
  • A – IV; B – III; C – V; D – I
  • A – II; B – III; C – IV; D – I
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Read List II from the other side.
Instead of matching each crystal system A, B, C, D one by one, look at each condition in List II and ask which crystal system it belongs to. This checks the same mapping from the reverse direction.

Step 2: Statement I says all edges and all angles are unequal.
A crystal system with no equal edges and no equal angles, and none of the angles fixed at $90^{\circ}$, is triclinic. So statement I belongs to D (Triclinic).

Step 3: Statement II says two edges equal, third different, all angles $90^{\circ}$.
$a = b \neq c$ with $\alpha = \beta = \gamma = 90^{\circ}$ is the defining condition of the tetragonal system. So statement II belongs to C (Tetragonal).

Step 4: Statement III says all edges unequal, two angles $90^{\circ}$ and one angle not $90^{\circ}$.
This one tilted angle with the rest at $90^{\circ}$, and no edges equal, is the monoclinic condition. So statement III belongs to B (Monoclinic).

Step 5: Statement IV says all edges equal and all angles $90^{\circ}$.
That is the most symmetric case, which is the cubic system. So statement IV belongs to A (Cubic).

Step 6: Put the matches together.
$A \to IV$, $B \to III$, $C \to II$, $D \to I$.
This is the same pairing as reading it forward, which confirms the answer. \[ \boxed{A-IV,\ B-III,\ C-II,\ D-I} \]
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