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Mekhanik [1.2K]
3 years ago
10

What atomic or hybrid orbitals make up the sigma bond between Cl and F in chlorine trifluoride, ClF3

Chemistry
1 answer:
Umnica [9.8K]3 years ago
7 0

Answer:

sp³d¹ hybridization

Explanation:

Given Cl as central element with three F substrates ...

The VSEPR structure indicates 5 hybrid orbitals that contain 2 diamagnetic orbitals (non-bonded e⁻-pairs) and 3 paramagnetic orbitals (single, non-paired electron for covalent bonding with fluorine) giving a trigonal bypyrimidal parent with a T-shaped geometry.

Valence bond theory predicts the following during bonding:

Cl:[Ne]3s²3p²p²p¹3d⁰

=> [Ne]3s²p²p¹p¹d¹

=> [Ne]3(sp³d)²(sp³d)²(sp³d)¹(sp³d)¹(sp³d)¹

giving 3 ( [Cl](sp³d) - [F]2p¹ ) sigma bonds and 2 non-bonded pairs on Cl.

Note the following images:

Non-bonded electron pairs are in plane of parent geometry and Fluorides covalently bonded to central element Chloride forming the T-shaped geometry.

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Th e molar absorption coeffi cients of tryptophan and tyrosine at 240 nm are 2.00 × 103 dm3 mol−1 cm−1 and 1.12 × 104 dm3 mol−1
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Answer:

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2.58x10⁻⁵ dm³mol⁻¹= Concentration of tryptophan

Explanation:

Lambert-Beer law is:

A = E×C×l

Where A is measured absorbance, E is absorption coefficient, C is concentration of solution and l is optical path length.

For the result at 240nm, it is possible to write:

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At 280 nm:

0.221 = 5.40x10³ dm³mol⁻¹cm⁻¹× Ctry × 1cm + 1.50x10³ dm³mol⁻¹cm⁻¹× Ctyr × 1cm <em>(2)</em>

<em></em>

Thus:

-2.7× (0.660 = 2.00x10³ × Ctry  + 1.12x10⁴ × Ctyr) =

-1.782 = -5.40x10³Ctry - 3.024x10⁴Ctyr

0.221 = 5.40x10³ × Ctry + 1.50x10³  × Ctyr

-------------------------------------------------------------------

-1.561 = -28740×Ctyr

<em>5.43x10⁻⁵ dm³mol⁻¹= Ctyr</em>

Replacing this value in (1) or (2) it is possible to find Ctry, that is:

<em>2.58x10⁻⁵ dm³mol⁻¹= Ctry</em>

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Explanation:

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