λ = h/mv

Butan-2-amine is 100% correct answer

Chlorophyll - Contains Mg complexed

Hemoglobin - Contains Fe complexed

Vitamin B₁₂ - Contains Co complexed

all organometalic compounds are formed by donation of pair of electron from ligands to central atom and they formed C --> M bonds which are coordinate in nature

few more examples are given below

diffraction experment as shown below

In multielectron atoms, the outermost electrons are shielded or screened from the nucleus by the inner electrons. This is known as shielding or screening effect. The outer most electrons do not feel the complete charge of the nucleus and the actual charge felt is called the effective nuclear charge. When the inner electrons are more, the screening effect will be large, the nuclear attraction will be less. Thus when the inner electrons increase the ionization energy will decrease.

it is given by deBroglies equation

de Broglie's suggested that all material objects including an electron have a dual character; they behave as particles as well as waves. The wavelength associated with a particle of mass 'm', moving with velocity 'v' is given by de Broglie's relation as:

λ = h/mv

Heisenberg's uncertainity principle it is not possible to measure simultaneously both the momentum (or velocity) and the position of a microscopic particle with absolute accuracy.

delta x * delta p greater than or equal to h/4pi

Mathematically this may be expressed delta p = uncertainity in momentum The constant on the right side of the equation (the product of the two uncertainties) tells us that the two uncertainties are inversely related. If the momentum of the particle is measured with more accuracy there will be a large uncertainity in its position and vice versa. Uncertainity is not due to the lack of refined techniques available, but because we cannot observe microscopic bodies without disturbing them. [Observations made as result of the impact of light suffer a change in the position or velocity of these microscopic objects]. This does not hold good for large objects of daily light, as the changes that occur are negligible.

Mole is defined as the amount of a substance, which contains the same number of chemical units (atoms, molecules, ions or electrons) as there are atoms in exactly 12 grams of pure carbon-12. A mole represents a collection of 6.023 x 10²³ ( Avogadro's number) chemical units. Thus a mole represents the quantity of material, which contains one Avogadro's number of chemical units of any substance. The unit of mole is denoted as 'mol'.

For example

One mole of hydrogen atoms = 6.023 x 10²³atoms of hydrogen

One mole of hydrogen molecules = 6.023 x 10²³ molecule of hydrogen

One mole of electrons = 6.023 x 10²³ electrons

One mole of sodium ions (Na⁺) = 6.023 x 10²³Na⁺ ions

It can be thus concluded that,

One mole of atoms = 6.023 x 10²³atoms = Gram atomic mass of the element.

One mole of molecules = 6.023 x 10²³molecules = Gram molecular mass.

the question is incomplete. please clarify with which they form pi complex for a proper nswer

Cl-O-Ca-Cl

the molecule has a bent shape like water

When metal oxides get reduced by carbon,one of the following process occur

MO + C ?? M + CO

MO + ½ C ?? M + ½ CO2

These equilibria can be discussed in terms of the thermodynamic function of the reactions,

1. M + ½ O2 ?? MO

2. ½ C + ½ O2 ?? ½ CO2

3. C + ½ O2 ?? CO

4. CO + ½ O2 ?? CO2

Temperature dependence of these reactions depend on the entropy change,

(dΔG/dT)P = −ΔS ΔS° of reaction 3 is higher than that of 2, since there is an increase in number of moles.In entropy units, the former is only 0.1 and the later is 8.5.Thus, ΔG decreases sharply with temperature for 3.

The Gibbs function for reaction 1 represents the metal’s affinity for oxygen. At room temperature ΔH° dominates ΔG°. The entropy of rn is about the same for different metals because of similar volume change.Thus ΔG has similar temperature dependence for different metals. Look at the graphs.

The kinks correspond to evaporation of the metals.Reduction of oxide by carbon depends on the affinity of metal to oxygen in comparison to carbon. Gibbs function for the relevant processes can be expressed in terms of the Gibbs functions for the oxidation rns. Inverting 1 and adding to 3

MO + C óM + CO ΔG° = ΔG°(3) - ΔG° (1)

MO + ½ Có M + ½ CO2 ΔG° = ΔGo(2) - ΔG° (1)

MO + CO ó M + CO2 ΔG° = ΔGo(4) - ΔGo(1)

Equilibrium lies to the right if ΔG° < 0.This will happen if ΔG° (1) lies below the carbon rns 2 – 4.

At any temperature the feasibility of the rn can be predicted examining the diagram. CuO can be reduced to copper any temperature above room temperature. Ag2O can be decomposed above 200°C simply by heating. Above 200°C the decomposition is spontaneous. Al2O3 can be ecomposed only above 2000°C. It cannot be reduced to metal by CO even up to 3000°C.Position of equilibrium at any temperature can be obtained by measuring the vertical separation between the lines.

Similar curves can be drawn for sulphides, nitrides, phosphates, chlorides etc.

Drawbacks of Ellingham diagram

1. Determination of minor energy differences becomes difficult if the lines lie closely.

2. ΔG° values do not take into account of activities which may be different form unity.

3. No account of the kinetics of the reaction is taken in measurements of thermodynamic quantities. Any oxide having a ΔG° value more negative than the given oxide can be used to reduce the oxide

When metal oxides get reduced by carbon,

one of the following process occur

MO + C <=>

M + CO M + ½ CO2

These equilibria can be discussed in terms of the

thermodynamic function of the reactions,

1. M + ½ O

2. ½ C + ½ O

2 <=> MO2 <=> ½ CO2
3. C + ½ O
4. CO + ½ O

Temperature dependence of these reactions depend on the
entropy change,
(d
that of 2, since there is an increase in number of moles.
In entropy units, the former is only 0.1 and the later is 8.5.
Thus,

ΔH° dominates
Δ
metals because of similar volume change.
Thus
metals. Look at the graphs.
The kinks correspond to evaporation of the metals.
Reduction of oxide by carbon depends on the affinity of
metal to oxygen in comparison to carbon. Gibbs function
for the relevant processes can be expressed in terms of the ns. Inverting 1 and
Gibbs functions for the oxidation r
adding to 3
G°. The entropy of rn is about the same for differentΔG has similar temperature dependence for different MO + C
MO + ½ C
MO + CO
??M + CO ΔG° = ΔG°(3) - ΔG° (1)?? M + ½ CO2 ΔG° = ΔGo(2) - ΔG° (1)?? M + CO2 ΔG° = ΔGo(4) - ΔGo(1)
Equilibrium lies to the right if
This will happen if
At any temperature the feasibility of the r
examining the diagram. CuO can be reduced to copper any
temperature above room temperature. Ag
above 200
spontaneous. Al
It cannot be reduced to metal by CO even up to 3000
Position of equilibrium at any temperature can be obtained by
measuring the vertical separation between the lines.
Similar curves can be drawn for sulphides, nitrides, phosphates,halides etc.
1
difficult if the lines lie closely.
2.
which may be different form unity.
3. No account of the kinetics of the reaction is taken in
measurements of thermodynamic quantities.
Any oxide having a
given oxide can be used to reduce the oxide
. Determination of minor energy differences becomesΔG° values do not take into account of activitiesΔG° value more negative than the
Drawbacks of Ellingham diagram
ΔG° < 0.ΔG° (1) lies below the carbon rns 2 – 4.n can be predicted2O can be decomposed°C simply by heating. Above 200°C the decomposition is2O3 can be decomposed only above 2000°C.°C.

The Gibbs function for reaction 1 represents the metal’s
affinity for oxygen. At room temperature
ΔG/dT)P = −ΔS ΔS° of reaction 3 is higher thanΔG decreases sharply with temperature for 3.
2 <=> CO2 <=> CO₂

MO + ½ C <=>

the general form of the hydroboration of alkenes mechanism is as follows:

First step is the attack of the alkene on BH₃, which then forms a four membered ring intermediate of partial bonds. It is because of this intermediate that hydroboration forms the anti-Markovnikov product. The boron atom is highly electrophilic because of its empty p orbital (ie. it wants electrons), and forms a slight bonding interaction with the pi bond. Since some electron density from the double bond is going towards bonding with the boron, the carbon opposite the boron is slightly electron deficient, left with a slightly positive charge. Positive charges are best stabilized by more highly substituted carbons, so the carbon opposite the boron tends to be the most highly substituted. Once the transition state breaks down, BH2 is attached to the least substituted carbon.

Peroxide then removes the borane and replaces it with the alcohol to form the anti-markovnikov product.

An example of the hydroboration mechanism:

Mathematically, the formation of
molecular orbitals may be described by the
linear combination of atomic orbitals that can
take place by addition and by subtraction of
wave functions of individual atomic orbitals
as shown below :
yMO = yA + yB
Therefore, the two molecular orbitals
s and s* are formed as :
s = yA + yB
s* = yA – yB
The molecular orbital s formed by the
addition of atomic orbitals is called the
bonding molecular orbital while the
molecular orbital s* formed by the subtraction
of atomic orbital is called antibonding
molecular orbital

y = Psi

s = sigma

s^* = sigma star

name and number longest carbon chain
identify branches
lowest number combinations
name each branch (alkyl group)

methyl CH₃

ethyl C₂H₅

propyl C₃H₇
For more than 1 of the same alkyl group use:

di = 2

tri = 3

tetra = 4
commas between numbers
hyphens between numbers and words
arrange branches alphabetically

Of course Mg is correct answer because all other elements are d block elements.

Magnesium is extracted from its ores by one of two processes. In the first, the ore is converted to magnesium chloride (MgCl₂), which is then electrolyzed. In the second process, the ore is converted to magnesium oxide (MgO), which is then treated with the alloy ferrosilicon. The ferrosilicon reacts with magnesium oxide to yield pure magnesium metal

Na can not be obtained as it has very high m.pt. and it becomes highly volatile at high temperature and produces toxic gases

Fused calcium chloride will cause more cooling because when it is added to Ice It causes depression in freezing point. which reduces the freezing point and it causes more cooling

BOND ORDER IN CASE OF CO IS 3 where as Bond order in case of CO₂ is 2

. As bond length decreases bond order increases

The C-O bond length in case of Carbon Monoxide is 112.8 pm where as in case of CO₂ it is 116 pm

The Boyle temperature is defined as the temperature for which the second virial coefficient, $B 2 (T)$ vanishes, i.e. $B 2 (T) = 0$ . Since higher order virial coefficients are generally much smaller than the second coefficient, the gas tends to behave as an ideal gas over a wider range of pressures when the temperature reaches the Boyle temperature. In any case, when the pressures are low, the second virial coefficient will be the only relevant one because the remaining concern terms of higher order on the pressure. We then have $d Z / d p = 0$ at $p = 0$ , where Z is the compression factor.

.

The coefficient, B(T), is a function of temperature and is called the "second virial coefficient. C(T) is called the third virial coefficient, and so on. The expansion is, in principle, an infinite series, and as such should be valid for all isotropic substances. In practice, however, terms above the third virial coefficient are rarely used in chemical thermodynamics.

Notice that we have set the quantity pV/nRT equal to Z. This quantity (Z) is called the "compression factor." It is a useful measure of the deviation of a real gas from an ideal gas. For an ideal gas the compression factor is equal to 1.

λ = h/mv

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