Match Column -1 molecule with Column - II shape- -Column-I - - Column-IINa- P4 - p- Crownllb - S8 - q - Polymeric dimeric c - AlCl3 - r - Tetrahedrallld - PCl - 5 - s - Trigonal bipyramidalll lendarray A- a r-b p- c q-d sB- a s-b p-c q-d rC- a r-b q-c p-d sD- a r-b p-c s-d q

The correct option is A (a) r; nbsp; nbsp; nbsp; nbsp; nbsp; (b) p; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; (c) q; ...

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Electric Potential and Potential Difference

Match Column ...

Question

Match Column -1 (molecule) with Column - II (shape):
$\begin{matrix} C o l u m n - I & C o l u m n - I I (a) & P_{4} & (p) & C r o w n (b) & S_{8} & (q) & P o l y m e r i c (d i m e r i c) (c) & A l C l_{3} & (r) & T e t r a h e d r a l (d) & P C l_{5} & (s) & T r i g o n a l b i p y r a m i d a l \end{matrix}$

(a) r; (b) p; (c) q; (d) s

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(a) s; (b) p; (c) q; (d) r

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(a) r; (b) q; (c) p; (d) s

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(a) r; (b) p; (c) s; (d) q

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Solution

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Column I

Column II

\begin{matrix} C o l u m n I & C o l u m n I I (T y p e o f △ A B C) a. cot \frac{A}{2} = \frac{b + c}{a} & p. always right angled b. a tan A + b tan B = (a + b) tan \frac{A + B}{2} & q. always isosceles c. a cos A = b cos B & r. may be right angled d. cos A = \frac{sin B}{2 sin C} & s. may be right angled isosceles \end{matrix}

\begin{matrix} Column -I & Column -II (a) N H_{4} C l & (P) Hydrogen bond (b) C u S O_{4} .5 H_{2} O & (Q) Co-ordinate bond (c) H N C & (R) Ionic bond (d) L i q u i d H_{2} O_{2} & (S) Covalent bond \end{matrix}

\begin{matrix} Column -I & Column -II (a) N H_{4} C l & (P) Hydrogen bond (b) C u S O_{4} .5 H_{2} O & (Q) Coordinate bond (c) H N C & (R) Ionic bond (d) L i q u i d H_{2} O_{2} & (S) Covalent bond \end{matrix}

s p^{3}

B F_{3}

s p^{2}

2 x^{2} - 7 x - 5 = 0

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