Chemistry, asked by manas8863, 11 months ago

only small amount of evaporation of water produces large cooling effects because of its

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Answered by debtwenty12pe7hvl
0

Only small amount of evaporation of water produces large cooling effects because -

Each molecule in the liquid phase is attracted to the other molecules in the liquid. When a molecule changes phase ( evaporates) it has to break all those intermolecular forces so it can escape. Only the molecules that have sufficient kinetic energy can do this. These high energy molecules leave to the gas phase and the remaining molecules are the ones that did not have enough energy to evaporate. This, the average kinetic energy of the molecules in the liquid left behind is lower that it was before the hot ones leave. So, evaporation “cools” what is left behind because the heat was used up breaking the hotter molecules out into the gas phase.

On a hot day you sweat. That sweat evaporates and carries away heat. Your body gave that heat to the sweat ( water) so your body cooled off. This is important as a means by which your body can avoid overheating.

Answered by mamidisettisanthosh2
0


Magnetic moments are often used in conjunction with electronic spectra to gain information about the oxidation number and stereochemistry of the central metal ion in coordination complexes.  

A common laboratory procedure for the determination of the magnetic moment for a complex is the Gouy method which involves weighing a sample of the complex in the presence and absence of a magnetic field and observing the difference in weight. A template is provided for the calculations involved.  

For first row transition metal ions in the free ion state, i.e. isolated ions in a vacuum, all 5 of the 3d orbitals are degenerate.  

A simple crystal field theory approach to the bonding in these ions assumes that when they form octahedral complexes, the energy of the d orbitals are no longer degenerate but are split such that two orbitals, the dx2-y2 and the dz2 (eg subset) are at higher energy than the dxy, dxz, dyz orbitals (the t2g subset).  

For octahedral ions with between 4 and 7 d electrons, this gives rise to 2 possible arrangements called either high spin/weak field or low spin/strong field respectively. The energy gap is dependent on the position of the coordinated ligands in the SPECTROCHEMICAL SERIES.

 

Note: For CHEM1902, we assume that all Co(III), d6 complexes are octahedral and LOW spin, i.e. t2g6.  

In tetrahedral complexes, the energy levels of the orbitals are again split, such that the energy of two orbitals, the dx2-y2 and the dz2 (e subset) are now at lower energy (more favoured) than the remaining three dxy, dxz, dyz (the t2 subset) which are destabilised.  

Tetrahedral complexes are ALL high spin since the difference between the 2 subsets of energies of the orbitals is much smaller than is found in octahedral complexes.  

The usual relationship quoted between them is: Δtet ≈ 4/9 Δoct.  

Square planar complexes are less commmon than tetrahedral and for CHEM1902 we will assume that the only ions forming square planar complexes are d8 e.g. Ni(II), Pd(II), Pt(II), etc. d8 can therefore be either square planar or tetrahedral. As with octahedral complexes, the energy gap between the dxy and dx2-y2 is Δoct and these d8 systems are all considered strong field / low spin complexes hence they are all diamagnetic, μ=0 Bohr Magneton (B.M.).

The formula used to calculate the spin-only magnetic moment can be written in two forms; the first based on the number of unpaired electrons, n, and the second based on the total electron spin quantum number, S. Since for each unpaired electron, n=1 and S=1/2 then the two formulae are clearly related and the answer obtained must be identical.

μso= √n(n+2) B.M.  

μso= √4S(S+1) B.M. - a variation of this will be introduced in the  

second year Inorganic course: μS+L = √{4S(S+1) + L(L+1)} B.M.

Comparison of calculated spin-only magnetic moments with experimentally observed data for some octahedral complexes

Ion Config μso / B.M. μobs / B.M.

Ti(III) d1 (t2g1) √3 = 1.73 1.6-1.7

V(III) d2 (t2g2) √8 = 2.83 2.7-2.9

Cr(III) d3 (t2g3) √15 = 3.88 3.7-3.9

Cr(II) d4 high spin (t2g3 eg1) √24 = 4.90 4.7-4.9

Cr(II) d4 low spin (t2g4) √8 = 2.83 3.2-3.3

Mn(II)/ Fe(III) d5 high spin (t2g3 eg2) √35 = 5.92 5.6-6.1

Mn(II)/ Fe(III) d5 low spin (t2g5) √3 = 1.73 1.8-2.1

Fe(II) d6 high spin (t2g4 eg2) √24 = 4.90 5.1-5.7

Co(III) d6 low spin (t2g6) 0 0

Co(II) d7 high spin (t2g5 eg2) √15 = 3.88 4.3-5.2

Co(II) d7 low spin (t2g6 eg1) √3 = 1.73 1.8

Ni(II) d8 (t2g6 eg2) √8 = 2.83 2.9-3.3

Cu(II) d9 (t2g6 eg3) √3 = 1.73 1.7-2.2

Comparison of calculated spin-only magnetic moments with experimental data for some tetahedral complexes

Ion Config μso / B.M. μobs / B.M.

Cr(V) d1 (e1) √3 = 1.73 1.7-1.8

Cr(IV) / Mn(V) d2 (e2) √8 = 2.83 2.6 - 2.8

Fe(V) d3 (e2 t21) √15 = 3.88 3.6-3.7

- d4 (e2 t22) √24 = 4.90 -

Mn(II) d5 (e2 t23) √35 = 5.92 5.9-6.2

Fe(II) d6 (e3 t23) √24 = 4.90 5.3-5.5

Co(II) d7 (e4 t23) √15 = 3.88 4.2-4.8

Ni(II) d8 (e4 t24) √8 = 2.83 3.7-4.0

Cu(II) d9 (e4 t25) √3 = 1.73 -


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