Home
Class 10
CHEMISTRY
A molecule of ammonia (NH(3)) has...

A molecule of ammonia `(NH_(3))` has

A

only single bonds

B

only double bonds

C

only triple bonds

D

two double bonds and one single bond

Text Solution

Verified by Experts

Promotional Banner

Topper's Solved these Questions

  • ATOMS AND MOLECULES

    VGS PUBLICATION-BRILLIANT|Exercise OBJECTIVE TYPE QUESTION|87 Videos

  • CHEMICAL BONDING

    VGS PUBLICATION-BRILLIANT|Exercise EXERCISE|302 Videos

Similar Questions

Explore conceptually related problems

When sodium (Na) metal is dissolved in liquid ammonia ("NH"_(3)) , it imparts a blue colour to the solution. This blue colouration is due to

(A) Ammonia is least volatile among MH_(3) type hydrides of group VA elements. (R ) Molecules of ammonia are held together by strong electrostatic forces

Calculate the molality of a solution by dissolving 0.850g of ammonia ( NH_3 ) in 100g of water.

The dissolution of ammonia gas in water does not obey Henry's law. On dissolving. a major portion of ammonia molecules unite with H_(2),O to form NH_(4),OH molecules. A portion of the latter again dissociates into NH_(4)^(+) , and OH^(-) ions. In solution thercfore, we have NH_(3) , molecules, NH_(4),OH molecules and NH_(4)^(+) , ions and the following equilibrium exist: NH_(3(g)) (pressure P and concentration c) rarr NH_(3(l)) + H_(2),Orarr NH_(4),OH rarr NH_(4)^(+) rarr,+OH^(-) Let c_(1) ,mol/L of NH_(3) , pass in liquid state which on dissolution in water forms c_(2) mol/ L of NH_(4),OH. The solution contains c_(3) mol/L of NH_(4)^(+) , ions. Total concentration of ammonia, which can be determined by volumetric analysis is equal to:

The dissolution of ammonia gas in water does not obey Henry's law. On dissolving. a major portion of ammonia molecules unite with H_(2),O to form NH_(4),OH molecules. A portion of the latter again dissociates into NH_(4)^(+) , and OH^(-) ions. In solution thercfore, we have NH_(3) , molecules, NH_(4),OH molecules and NH_(4)^(+) , ions and the following equilibrium exist: NH_(3(g)) (pressure P and concentration c) rarr NH_(3(l)) + H_(2),Orarr NH_(4),OH rarr NH_(4)^(+) rarr,+OH^(-) Let c_(1) ,mol/L of NH_(3) , pass in liquid state which on dissolution in water forms c_(2) mol/ L of NH_(4),OH. The solution contains c_(3) mol/L of NH_(4)^(+) , ions. Degree of dissociation of ammonium hydroxide is

Compare diploment of NH_(3) molecule with that of NF_(3) molecule .

The dissolution of ammonia gas in water does not obey Henry. s law. On dissolving , a major portion of ammonia molecules unite with H_2O to form NH_4 OH molecules . A portion of the latter again dissociates into NH_4^(+) and OH ^(-) ions. In solution therefore , we have NH_3 molecules, NH_4OH molecules and NH_4^(+) ions and the following equilibrium exist : NH_(3(g) ) (pressure P and concentration c) hArr NH_(3(i)) +H_2O hArr NH_4 OH hArr NH_4^(+) +OH^(-) Let c_1 mol//L of NH_3 pass in liquid state which on dissolution in water forms c_2 mol/ L of NH_4 OH. The solution contains c_3 mol/L of NH_4^(+) ions. The dissociation constant of NH_4 OH can be given as :