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kolezko [41]
3 years ago
14

If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the struct

ure The central X atom is single bonded to two Y atoms, which are 180 degrees apart. The X atom has two lone pairs. could be abbreviated as XY2Z2. Classify each molecule according to its shape.Molecular Geometries: (A) linear(B) bent 120(C) bent 109(D) trigonal pyramidal(E) T-shaped(F) see saw(G) square planar(H) square pyramidalPossiblities:(1) XY4Z2(2) XY5Z(3) XY2Z(4) XY2Z3(5) XY4Z(6) XY2Z2(7) XY3Z2(8) XY3Z
Chemistry
1 answer:
sashaice [31]3 years ago
5 0

Answer:

(C) bent 109

(2) XY_5Z

Explanation:

The number of bond pairs = 2

Number of lone pairs = 2

The general formula is - XY_2Z_2

According to VSEPR theory, the atoms will form a geometry in such a way that there is minimum repulsion and maximum stability.  

The central atom has 2 bond pairs and 2 lone pairs. Hybridization is sp³. The geometry is bent shape which has an angle of approximately 105.5°. It original has angle of 109.5° but due to lone pair repulsion, it reduces.

Answer - (C) bent 109

Square pyramidal possibility:- (2) XY_5Z

The central atom has 5 bond pairs and 1 lone pair. Hybridization is sp³d. The geometry is square pyramidal in which the equatorial bonds has an angle of 120° and axial bond has an angle of 90°.

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The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

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