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When cell has stalled DNA replication fork, which checkpoint should be predominantly activated?

Option: 1

G1/S

 

Option: 2

G2/M

 

 

 

Option: 3

M

 

 

Option: 4

Both G2 M and M

 

G2/M should be activated as the cell has stalled DNA replication fork.

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Posted by

Ajit Kumar Dubey

 The group having triangular planar structures is :  
Option: 1BF_3, NF_3,CO_3^{2-}
Option: 2 CO ^{2-}_3, NO_3^-, SO_3
Option: 3 NH_3, SO_3,CO^{2-}_3  
Option: 4 NCl_3, BCl_3, SO_3  
 

The group having triangular planar structures is CO32-, NO?????3?-,SO?3
In CO32- ion, C atom is sp?????2 hybridised. This results in triangular planar structure.
In NO?????3????- ion, N atom is sp?????2 hybridised. This results in triangular planar structure.
In SO?????3, S atom is sp?????2 hybridised. This results in triangular planar structure.
In all above molecules/ions, the central atom has 3 bonding domains and bond angle of 120o. 

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Posted by

vishal kumar

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 The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm.  This transition is associated with :
Option: 1  Lyman series
Option: 2  Balmer series
Option: 3  Paschen series
Option: 4  Brackett series  
 

For hydrogen atom, the radius of n??????th orbit, r??????n = 52.9 pm ×n2

211.6 pm = 52.9 pm×n2

n??????2=4

n=2

Hence, the transition is from a higher orbit to the second orbit.
This corresponds to the Balmer series.

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Posted by

vishal kumar

 If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of He+ is :  
Option: 1 \frac{5A}{9}
Option: 2 \frac{9A}{5}
Option: 3 \frac{36A}{5}
Option: 4 \frac{36A}{7}

we know that,

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vishal kumar

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For the Balmer series in the spectrum of H atom, \bar{v}=\mathrm{R}_{\mathrm{H}}\left\{\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right\} the correct statements among (I) to (IV) are: (1) As wavelength decreases, the lines in the series converge
(II) The integer n1 is equal to 2
(III) The lines of longest wavelength corresponds to n2 = 3
(IV) The ionization energy of hydrogen can be calculated from wavenumber of these lines
 
Option: 1 (II),(III),(IV)
Option: 2 (I),(III),(IV)
Option: 3 (I),(II),(III)
Option: 4 (I),(II),(IV)
 

Line Spectrum of Hydrogen-like atoms

\frac{1}{\lambda }= RZ^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )

Where R is called Rydberg constant, R = 1.097 X 107,  Z is the atomic number

n1= 1, 2, 3….

n2= n1+1, n1+2 ……

Lyman Series spectrum:

Where

n1= 1 and  n2= 2, 3, 4....

This lies in the Ultraviolet region.

Balmer Series Spectrum:

Where n1= 2 and  n2= 3, 4, 5, 6....

It lies in the visible region.

Paschen, Bracket and Pfund Series spectrums:

these lies in Infrared Region.

-

Since, \mathrm{\Delta E\: \propto\:\frac{1}{\lambda} } so for wavelength to be longest or maximum the energy gap should be minimum. For the Balmer series, the transition for the longest wavelength is from n = 3 to n = 2. Thus clearly, n1 value for the Balmer series is 2. Further, as wavelength decreases, the lines in the series converge.

Therefore, Option(3) is correct.

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Posted by

vishal kumar

The number of orbitals associated with quantum numbers n = 5, m_s = +1/2
is:
Option: 1 25
Option: 2 11
Option: 3 15
Option: 4 50
 

We have:

n = 5, ms = +1/2

Thus, the values of l are from 0 to (n-1)

l = 0 to 4

Thus, values of l are 5s, 5p, 5d, 5f, and 5g

Now, the total number of orbitals = n^2 = 5^2 = 25

Therefore, Option(1) is correct.

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Posted by

Kuldeep Maurya

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The dipole moments of CCl4, CHCl3 and CH4 are in the order:
Option: 1 CH_4=CCl_4<CHCl_3
Option: 2 CHCl_3<CH_4=CCl_4
Option: 3 CH_4<CCl_4<CHCl_3
Option: 4 CH_4<CCl_4<CHCl_3
 

Let us first look at the structures of the given compounds

\mathrm{Clearly \ \mu \neq 0 \textrm{ of CHCl}_3}

Therefore, Option(1) is correct.

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Posted by

Kuldeep Maurya

The theory that can completely /properly explain the nature of bonding in[Ni(CO)_4]
Option: 1 Molecular orbital theory
 
Option: 2 Crystal field theory
 
Option: 3 Wemer's theory  
 
Option: 4 Valence bond theory
 
 

As we have learnt,

 

Molecular Orbital Theory -

Molecular orbital theory (MO theory) provides an explanation of chemical bonding that accounts for the paramagnetism of the oxygen molecule. It also explains the bonding in a number of other molecules, such as violations of the octet rule and more molecules with more complicated bonding that are difficult to describe with Lewis structures. Additionally, it provides a model for describing the energies of electrons in a molecule and the probable location of these electrons. Unlike valence bond theory, which uses hybrid orbitals that are assigned to one specific atom, MO theory uses the combination of atomic orbitals to yield molecular orbitals that are delocalized over the entire molecule rather than being localized on its constituent atoms. MO theory also helps us understand why some substances are electrical conductors, others are semiconductors, and still others are insulators. The table given below explains the major differences between the valence bond theory and molecular orbital theory.

 

Comparison of Bonding Theories

Valence Bond Theory

Molecular Orbital Theory

considers bonds as localized between one pair of atoms

considers electrons delocalized throughout the entire molecule

creates bonds from overlap of atomic orbitals (s, p, d…) and hybrid orbitals (sp, sp2, sp3…)

combines atomic orbitals to form molecular orbitals (σ, σ*, π, π*)

forms σ or π bonds

creates bonding and antibonding interactions based on which orbitals are filled

predicts molecular shape based on the number of regions of electron density

predicts the arrangement of electrons in molecules

needs multiple structures to describe resonance

  

 

Molecular orbital theory describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, Ψ, analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a molecular orbital (Ψ2). Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin.

We will consider the molecular orbitals in molecules composed of two identical atoms (H2 or Cl2, for example). Such molecules are called homonuclear diatomic molecules. In these diatomic molecules, several types of molecular orbitals occur.

The mathematical process of combining atomic orbitals to generate molecular orbitals is called the linear combination of atomic orbitals (LCAO). The wave function describes the wavelike properties of an electron. Molecular orbitals are combinations of atomic orbital wave functions. Combining waves can lead to constructive interference, in which peaks line up with peaks, or destructive interference, in which peaks line up with troughs as shown in the figure below. In orbitals, the waves are three dimensional, and they combine with in-phase waves producing regions with a higher probability of electron density and out-of-phase waves producing nodes, or regions of no electron density.

A pair of diagrams are shown and labeled, “a” and “b.” Diagram a shows two identical waves with two crests and two troughs. They are drawn one above the other with a plus sign in between and an equal sign to the right. To the right of the equal sign is a much taller wave with a same number of troughs and crests. Diagram b shows two waves with two crests and two troughs, but they are mirror images of one another rotated over a horizontal axis. They are drawn one above the other with a plus sign in between and an equal sign to the right. To the right of the equal sign is a flat line.

(a) When in-phase waves combine, constructive interference produces a wave with greater amplitude. (b) When out-of-phase waves combine, destructive interference produces a wave with less (or no) amplitude.

 

-

 

 

The metal-carbonyl \pi is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding \pi* orbital of carbon monoxide.

Therefore, Option(1) is correct.

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Posted by

Kuldeep Maurya

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Complexes (ML_{5}) of metals Ni and Fe have ideal square pyramidal and trigonal bipyramidal geometries, respectively.The sum of the 90^{\circ},120^{\circ} and 180^{\circ}L-M-L angles in the two complexes is _________.
 

As we have learnt,

 

Shapes of Molecules -

The ideal shapes of molecules, which are predicted on the basis of electron pairs and lone pairs of electrons are mentioned in the table below:

A table is shown that is comprised of six rows and six columns. The header row reads: “Number of Electron Pairs,” “Electron pair geometries; 0 lone pair,” “1 lone pair,” “2 lone pairs,” “3 lone pairs,” and “4 lone pairs.” The first column contains the numbers 2, 3, 4, 5, and 6. The first space in the second column contains a structure in which the letter E is single bonded to the letter X on each side. The angle of the bonds is labeled with a curved, double headed arrow and the value, “180 degrees.” The structure is labeled, “Linear.” The second space in the second column contains a structure in which the letter E is single bonded to the letter X on three sides. The angle between the bonds is labeled with a curved, double headed arrow and the value, “120 degrees.” The structure is labeled, “Trigonal planar.” The third space in the second column contains a structure in which the letter E is single bonded to the letter X four times. The angle between the bonds is labeled with a curved, double headed arrow and the value, “109 degrees.” The structure is labeled, “Tetrahedral.” The fourth space in the second column contains a structure in which the letter E is single bonded to the letter X on five sides. The angle between the bonds is labeled with a curved, double headed arrow and the values “90 and 120 degrees.” The structure is labeled, “Trigonal bipyramid.” The fifth space in the second column contains a structure in which the letter E is single bonded to the letter X on six sides. The angle between the bonds is labeled with a curved, double headed arrow and the value, “90 degrees.” The structure is labeled, “Octahedral.” The first space in the third column is empty while the second contains a structure in which the letter E is single bonded to the letter X on each side and has a lone pair of electrons. The angle between the bonds is labeled with a curved, double headed arrow and the value, “less than 120 degrees.” The structure is labeled, “Bent or angular.” The third space in the third column contains a structure in which the letter E is single bonded to the letter X three times and to a lone pair of electrons. It is labeled with a curved, double headed arrow and the value, “less than 109 degrees.” The structure is labeled, “Trigonal pyramid.” The fourth space in the third column contains a structure in which the letter E is single bonded to the letter X on four sides and has a lone pair of electrons. The bond angle is labeled with a curved, double headed arrow and the values, “less than 90 and less than 120 degrees.” The structure is labeled, “Sawhorse or seesaw.” The fifth space in the third column contains a structure in which the letter E is single bonded to the letter X on five sides and has a lone pair of electrons. The bond angle is labeled with a curved, double headed arrow and the value, “less than 90 degrees.” The structure is labeled, “Square pyramidal.” The first and second spaces in the fourth column are empty while the third contains a structure in which the letter E is single bonded to the letter X on each side and has two lone pairs of electrons. The bond angle is labeled with a curved, double headed arrow and the value, “less than less than 109 degrees.” The structure is labeled, “Bent or angular.” The fourth space in the fourth column contains a structure in which the letter E is single bonded to the letter X three times and to two lone pairs of electrons. The bond angle is labeled with a curved, double headed arrow and the value, “less than 90 degrees.” The structure is labeled, “T - shape.” The fifth space in the fourth column contains a structure in which the letter E is single bonded to the letter X on four sides and has two lone pairs of electrons. The bond angle is labeled with a curved, double headed arrow and the value “90 degrees.” The structure is labeled, “Square planar.” The first, second and third spaces in the fifth column are empty while the fourth contains a structure in which the letter E is single bonded to the letter X on each side and has three lone pairs of electrons. The bond angle is labeled with a curved, double headed arrow and the value, “180 degrees.” The structure is labeled, “Linear.” The fifth space in the fifth column contains a structure in which the letter E is single bonded to the letter X three times and to three lone pairs of electrons. The bond angle is labeled with a curved, double headed arrow and the value, “less than 90 degrees.” The structure is labeled, “T - shape.” The first, second, third, and fourth spaces in the sixth column are empty while the fifth contains a structure in which the letter E is single bonded to the letter X on each side and has four lone pairs of electrons. The bond angle is labeled with a curved, double headed arrow and the value “180 degrees.” The structure is labeled, “Linear.” All the structures use wedges and dashes to give them three dimensional appearances.

 

Predicting the geometry of molecules 

The following procedure uses VSEPR theory to determine the geometry of the molecules:

  1. Write the Lewis structure of the molecule or polyatomic ion.

  2. Count the number of regions of electron density (lone pairs and bonds) around the central atom. A single, double, or triple bond counts as one region of electron density.

  3. Identify the electron-pair geometry based on the number of regions of electron density: linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral 

  4. Use the number of lone pairs to determine the molecular structure. If more than one arrangement of lone pairs and chemical bonds is possible, choose the one that will minimize repulsions, remembering that lone pairs occupy more space than multiple bonds, which occupy more space than single bonds. In trigonal bipyramidal arrangements, repulsion is minimized when every lone pair is in an equatorial position. In an octahedral arrangement with two lone pairs, repulsion is minimized when the lone pairs are on opposite sides of the central atom.

For example, BCl3 has three electron pairs and no lone pairs of electrons. Thus these three electron pairs will arrange themselves in a trigonal planar geometry as shown below. The bond angles between each B-Cl bonds is 1200.

 

-

   \angle 120^{\circ}=3;\angle 90^{\circ}=6;\angle 180^{\circ}=1 \Rightarrow Total=10

  \angle 90^{\circ}=8;\angle 180^{\circ}=2 \Rightarrow Total=10

Thus, the correct answer is 20.

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Posted by

Kuldeep Maurya

The radius of the second Bohr orbit, in terms of the Bohr radius, a_{0}, in Li^{2+} is : 
Option: 1 \frac{2a_{0}}{3}

Option: 2 \frac{4a_{0}}{3}

Option: 3 \frac{4a_{0}}{9}

Option: 4 \frac{2a_{0}}{9}
 

As we have learnt,

 

Radius, velocity and the energy of nth Bohr orbital -

 

Bohr radius of nth orbit:

r_{n}= 0.529 \frac{n^{2}}{z}A^{0}

where Z is the atomic number

Velocity of electron in nth orbit:

v_{n}= (2.165\times 10^{6})\frac{z}{n}\: m/s

where z is atomic number

Total energy of electron in nth orbit:

E_{n}= -13.6\: \frac{z^{2}}{n^{2}}eV

Where z is atomic number

-

 

r=\frac{n^2a_o}{Z}.

\mathrm{For\ Li^{2+}\ r=\frac{2^2a_o}{3}=\frac{4a_o}{3}}

Therefore, Option(2) is correct.

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Posted by

vishal kumar

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