a)HCl consists of one H and Cl atom separated by 0.127 nm, the bond length. The

Question

a)HCl consists of one H and Cl atom separated by 0.127 nm, the bond length. The Cl atom has a partial charge of -0.177e and the H atom has a partial charge of +0.177e, where e = 1.602x10-19 C is the electron charge.What is the dipole moment of the molecule?

b)Now the HCl molecule is placed near a sodium ion Na+ with charge +1e. As shown in the figure, the distance between the Cl and Na atoms is d = 1.2 nm.What is the magnitude of the total force on the HCl molecule due to the Na+ ion |FHCl|?

c)Compare your previous answer to the force on a single Cl- ion with charge -1e due to the Na+ ion 1.2 nm away.

What is the ratio of the force on the Cl- ion to the force on the HCl molecule |FCl-/FHCl|?

Explanation / Answer

a)

q = charge of dipole = 0.177 e = 0.177 x 1.6 x 10-19 C

L = length of dipole = 0.127 nm = 0.127 x 10-9 m

Dipole moment is given as

P = qL = 0.177 x 1.6 x 10-19 (0.127 x 10-9 ) = 3.6 x 10-30 Cm

b)

Q = charge on Na = 1.6 x 10-19 C

Force on Cl = Fcl = k q Q / d2           towards left

Force on H = Fh = k q Q / (d + L)2        towards right

net force = Fnet = Fcl - Fh = k q Q / d2 - k q Q / (d + L)2  

Fnet = k q Q [1/d2 - 1/(d + L)2]

Fnet = (9 x 109) (0.177 x 1.6 x 10-19 ) (1.6 x 10-19 ) [1/(1.2 x 10-9)2 - 1/(1.2 x 10-9 + 0.127 x 10-9)2 ]

Fnet = 5.2 x 10-12 N

c)

Fcl = k q Q / d2   = (9 x 109) (0.177 x 1.6 x 10-19 ) (1.6 x 10-19 )/(1.2 x 10-9)2 = 2.83 x 10-11 N

Ratio = Fcl/Fhcl = 2.83 x 10-11 / (5.2 x 10-12 ) = 5.44

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