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Sunday, April 30, 2017

Puzzle 4


Puzzle 1: A man was outside taking a walk, when it started to rain. The man didn’t have an umbrella and he wasn’t wearing a hat. His clothes got soaked, yet not a single hair on his head got wet. How could this happen?

Puzzle 2: A cowboy rides into town on Friday, stays for three days and then leaves on Friday. How did he do it?

Puzzle 3: You walk across a bridge and you see a boat full of people yet there isn’t a single person on board. How is that possible?

Puzzle 4: A boy was rushed to the hospital emergency room. The ER doctor saw the boy and said, “I cannot operate on this boy. He is my son.” But the doctor was not the boy’s father. How could that be?

Puzzle 5: What can run but can’t walk?

Photo Electric Effect

It is the emission of electrons from the surface of certain substances, mainly metals, when they are illuminated by electromagnetic radiations like X-rays, ultraviolet and even visible light. The electrons emitted are called photo electrons.

Experiment to Demonstrate Photoelectric Effect:


In figure (a) A and C are two zinc plates, connected to positive and negative of a battery E respectively, through a micro-ammeter M and a key K. The plates A and C are enclosed in an evacuated quartz bulb. The quartz will not absorb ultraviolet rays. It is evacuated, so that, the metal surface remains pure.

The micro ammeter M registers a current, whenever ultraviolet rays fall on the zinc plate C. These electrons travel towards the plate A and thus a current flows through the circuit. If ultraviolet rays are stopped, there is no deflection in the micro ammeter. This means, electrons are ejected from the plate C by the radiations falling on it.

Laws of Photo Electric Emission:

⇒ For a given photosensitive material, there is a minimum frequency below which there is no photoelectric emission. This frequency is independent of the intensity of light and is called the threshold frequency.

⇒ Photoelectric emission is an instantaneous phenomenon. There is no time lag between the incidence of radiation and the emission of photo electrons.

⇒ The number of photo electrons and the photoelectric current is directly proportional to the intensity of incident radiation (provided the frequency  is greater than the threshold frequency).

⇒ The kinetic energy of photo electrons increases with the frequency of the incident radiation and is independent of the intensity of radiation.

Saturday, April 29, 2017

Brain Teaser 11 [With Answers]


Puzzle 1: 14 students in a class failed in both QT and Accounts. 20 failed in QT, but 32 failed in Accounts. How many students were there in total?

Answer: 20 + 32 - 14 = 38 students.

Puzzle 2: Solve this puzzle:

S is as much older than K as he is younger than P. N is as old as K .Which of the following is wrong?

a)      K is younger than P.

b)      N is younger than P.

c)      S is older than N.

d)      P is not the oldest.

e)      K is younger than S.

Answer: N = K < S < P. So d is wrong

Puzzle 3: If two's company, and three's a crowd, what are four and five?

Answer: Nine.

Puzzle 4: If you have it, you want to share it. If you share it, you don't have it. What is it?

Answer: A secret.

Puzzle 5: If you threw a White stone into the Red Sea, what would it become?

Answer: Wet

Standard form of the ellipse

Let S be the focus, ZK the directrix and e the eccentricity of the ellipse whose equation is required. 

Draw SK perpendicular from S on the directrix. Divide SK internally and externally at A and A’ (on SK produced) respectively in the ratio e : 1.

∴ ⇒ SA = e. AK … (1)

And
            
 ⇒ SA’ = e A’K … (2)

Since A and A’ are such points that their distances from the focus bear constant ratio e(< 1) to their respective distances from the directrix. Therefore these points lie on the ellipse.

Let AA’ = 2a and C be the mid - point of AA’.

Then CA = CA’ = a

Adding (i) and (ii), we get

SA + SA’ = e (AK + A’K)

2a = e (CK – CA + A’C + CK)

2a = 2e CK [ CA = A’C = a]

CK = a/e … (3)


Subtracting (i) from (ii), we get

SA’ - SA = e (A’K - AK)

(SC + CA’) – (CA - CS) = e (AA’)

2CS = 2ae

CS = ae

Now let us choose C as the origin. CAX as x-axis and a line are (ae, 0) and equation of the directrix ZK is z = a/e.

Let P(x, y) be any point on the ellipse. Join SP and draw PMZK. Then, by definition, we have

SP = e PM

SP² = e² PM²

x² (1 - e²) + y² = a² (1 - e²)

⇒ 

, Where b² = a² (1 - e²)

This is the standard equation of the ellipse.

Note:

e < 1 1 - e² < 1 a² (1 - e²) < a² b² = a².

Tracing of the ellipse , when a > b:

We have , where a > b … (1)

  … (2)

And  … (3)

We observe the following:

a. Symmetry: For every value of x there are equal and opposite values of y. Similarly for every value of y there are equal and opposite values of x.

Thus, the curve is symmetric about both the axes.

b. Origin: The curve does not pass through the origin.

c. Intersection with axes: The curve meets x axis at y = 0

Putting y = 0 (iii), we get x = ± a.

So the curve meets y-axis at A (a, 0) and A’ (-a, 0).

Putting x = 0 in (ii), we get y = ± b.

So the curve meets y-axis at B (0, b) and B’ (0, -b).

d. Region: if x > a or, x < -a, from (ii) we get imaginary values of y, therefore, there is no part of the curve to the right of A or to the left of A’.

If y > b or y < -b, from (iii) we get imaginary values of x. therefore, there is no part of the curve above B (0, b) or below B’ (0, -b).

From (ii), we find that at x = 0, y = ±b and as x increases the convenient points on the ellipse the general shape of the ellipse 

Friday, April 28, 2017

Brain Teaser 11


Teaser 1: 14 students in a class failed in both QT and Accounts. 20 failed in QT, but 32 failed in Accounts. How many students were there in total?

Teaser 2: Solve this puzzle:

S is as much older than K as he is younger than P. N is as old as K .Which of the following is wrong?

a) K is younger than P.

b)      N is younger than P.

c)      S is older than N.

d)      P is not the oldest.

e)      K is younger than S.

Teaser 3: If two's company, and three's a crowd, what are four and five?

Teaser 4: If you have it, you want to share it. If you share it, you don't have it. What is it?

Teaser 5: If you threw a White stone into the Red Sea, what would it become?

Crystal Field Theory (CFT)

Crystal field theory differs from valence bond theory in viewing the complex as held together by purely electrostatic attractions. That is, it ignores covalent bonding.

The most significant aspects of the theory is its concern with the effect that the ligands have on the energies of the d-orbitals of the metal. The ligands are viewed as pointing their negative end in the direction of the metal cation.

The theory is developed by considering energy changes of the five degenerate d-orbitals upon being surrounded by an array of point charges consisting of the ligands.

As a ligand approaches the metal ion, the electrons from the ligand will be closer to some of the d-orbitals and farther away from others causing a loss of degeneracy.

The electrons in the d-orbitals and those in the ligand repel each other due to repulsion between like charges. Thus the d-electrons closer to the ligands will have a higher energy than those further away which results in the d-orbitals splitting in energy.

This splitting is affected by the following factors:

a) The nature of the metal ion.

b) The metal's oxidation state. A higher oxidation state leads to a larger splitting.

c) The arrangement of the ligands around the metal ion.

d) The nature of the ligands surrounding the metal ion. The stronger the effect of the ligands then the greater the difference between the high and low energy d groups.

The most common type of complex is octahedral; here six ligands form an octahedron around the metal ion. In octahedral symmetry the d-orbitals split into two sets with an energy difference, Δoct (the crystal-field splitting parameter) where the dxy, dxz and dyz orbitals will be lower in energy than the dz2 and dx2-y2, which will have higher energy, because the former group is farther from the ligands than the latter and therefore experience less repulsion.

The three lower-energy orbitals are collectively referred to as t2g, and the two higher-energy orbitals as eg


 

The size of the gap Δ between the two or more sets of orbitals depends on several factors, including the ligands and geometry of the complex.

Some ligands always produce a small value of Δ, while others always give a large splitting. The reasons behind this can be explained by ligand field theory.

The spectrochemical series is an empirically-derived list of ligands ordered by the size of the splitting Δ that they produce:

I ⁻ < Br⁻ < S²⁻ < SCN⁻ < Cl⁻ < NO₃⁻ < N₃⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < CH₃CN < py < NH₃ < en < 2, 2’ - bipyridine < phen < NO₂⁻ < PPh₃ < CN⁻ < CO.

It is useful to note that the ligands producing the most splitting are those that can engage in metal to ligand back-bonding. For d4 ions 2 possibilities are there:

1. The fourth electron could enter t2g level and pair with an electron

2. It can enter eg level and avoid pairing

The selection of one of these possibilities depends on magnitude of splitting, ∆0 and pairing energy, P

1. If ∆0 < P, electron enters eg orbitals.
Ligands with ∆0 < P are weak field ligands and form high spin complexes

2. If ∆0 > P, electron enters t2g orbital.
Ligands with ∆0 > P are strong field ligands and form low spin complexes

Tetrahedral complexes are the second most common type; here four ligands form a tetrahedron around the metal ion.

In a tetrahedral crystal field splitting the d-orbitals again split into two groups, with an energy difference of Δtet where the lower energy orbitals will be dz2 and dx2-y2, and the higher energy orbitals will be dxy, dxz and dyz - opposite to the octahedral case.

Furthermore, since the ligand electrons in tetrahedral symmetry are not oriented directly towards the d-orbitals, the energy splitting will be lower than in the octahedral case. Square planar and other complex geometries can also be described by CFT.


Colour in coordination compounds:

The colour of the complex is complementary to that which is absorbed.

If a complex whose d orbitals are split absorbs a photon of visible light, one or more electrons may momentarily jump from the lower energy d-orbitals to the higher energy ones to transiently create an excited state atom.

The difference in energy between the atom in the ground state and in the excited state is equal to the energy of the absorbed photon.

Because only certain wavelengths (λ) of light are absorbed - those matching exactly the energy difference - the compounds appears the appropriate complementary color.

In the absence of ligand, crystal field splitting doesn’t occur and hence substance is colourless.
Ex: anhydrous CuSO4 is colourless but CuSO4.5H2O is blue in colour.

Coordination Entity
Wavelength of Light Absorbed (nm)
Colour of Light Absorbed
Colour of Coordination Entity
[CoCl(NH₃)₅]²⁺
535
Yellow
Violet
[Co(NH₃)₅(H₂O)]³⁺
500
Blue Green
Red
[Co(NH₃)₆]³⁺
475
Blue
Yellow Orange
[Co(CN)₆]³⁻
310
Ultraviolet
Pale Yellow
[Cu(H₂O)₄]²⁺
600
Red
Blue
[Ti(H₂O)₆]³⁺
498
Blue Green
Purple

Limitations of crystal field theory:

According to crystal field theory anionic ligands are point charges, then anionic ligands should act as strong ligands and should exert the greatest splitting effect. But, anionic ligands are actually found at the low end of the spectro-chemical series.

It does not take into account the covalent character of bonding between the ligand and the central atom.