Select the high spin complexes from the following:
(A) \(\left[ Ni ( CN )_4\right]^{2-}\)
(B) \(\left[ Fe ( CN )_6\right]^{3-}\)
(C) \(\left[ MnBr _4\right]^{2-}\)
(D) \(\left[ Fe \left( H _2 O \right)_6\right]^{2+}\)
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A first order reaction is found to have a rate constant, k = 5.5 × 10^-14 s^-1. The half life of the reaction will be :

Match the statement of set of orbitals given in List-I with orbitals and energy separation given in List-II

List – I (Statements)	List-II (Orbitals and energy separation)
(A) \(t_{2 g}\) set of orbitals in octahedral crystal field and energy separation from barycentre	(I) \(d_{x^2-y^2}, d_{z^2}\) and \(3 / 5 \Delta_t\)
(B) \(e_g\) set of orbitals in octahedral crystal field and energy separation from barycentre	(II) \(d_{x y}, d_{x z}, d_{y z}\) and \(2 / 5 \Delta_t\)
(C) \(t_2\) set of orbitals in tetrahedral crystal field and energy separation from barycentre	(III) \(d_{x^2-y^2}, d_{z^2}\) and \(3 / 5 \Delta_0\)
(D) \(e\) set of orbitals in tetrahedral crystal field and energy separation from barycentre	(IV) \(d_{x y}, d_{x z}, d_{y z}\) and \(2 / 5 \Delta_0\)

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Read the passage carefully and answer the Questions
The concentration dependence of the rate of a reaction is called the differential rate equation. It is not always convenient to determine the instantaneous rate, as it is measured by calculating the slope of the tangent at point 't' in the concentration vs time plot. This makes it difficult to determine the rate and hence the order of the reaction. This difficulty is overcome by integrating the differential rate equation. This integrated rate equation gives a direct relation between concentrations at different times and the rate constant. The integrated rate equations are different for the reactions having different reaction orders.
The rate constant of a reaction is 2.1 mol L-'s-¹. What is the order of the reaction?

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The degeneracy of the \(d\) orbitals has been removed due to ligand electron-metal electron repulsions in the octahedral comples to yield three orbitals of lower energy, \(t_{2 g}\) set and two orbitals of higher energy, \(e_g\) set. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting and the energy separation is denoted by \(\Delta_0\). Thus energy of the two \(e_g\) orbitals will increase by \((3 / 5) \Delta_0\) and that of three \(t_{2 g}\) will decrease by \((2 / 5) \Delta_0\). The crystal field splitting \(\Delta_0\) depends upon the field produced by the ligand and charge on the metal ion. Some ligands are able to produce strong fields, in which case the splitting will be large, whereas others produce weak fields and consequently result in small splitting of \(d\) orbitals. Relative magnitude of crystal field splitting energy \(\Delta_{\circ}\) and pairing energy, \(P\) (energy required for electron pairing in a single orbital) determine the formation of low spin ( \(\Delta_0>P\) ) or high spin ( \(\Delta_0<P\) ) complex. In tetrahedral coordination entity formation, the \(d\)-orbital splitting is inverted and is smaller as compared to the octahedral field splitting. The crystal field theory attributes the colour of the complex to \(d-d\) transition of the electron. The colour of the coordination compounds depends on the crystal field splitting. In the absence of ligand, crystal field splitting does not occur and hence the substance is colourless.
Choose the correct order of the ligands in their increasing crystal field splitting:

Which of the following is the best reducing agent to convert RCOOH into \(RCH _2 OH\) ?

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The transition elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example, exhibits all the oxidation states from +2 to +7 . The lesser number of oxidation states at extreme ends stems from either too few electrons to lose or share (Sc, Ti) on too many d-electrons (hence fewer orbitals available in which to share electrons with others) for higher valence \(( Cu , Zn )\). Thus early in the series scandium (II) is virtually unknown and titanium (IV) is more stable than Ti (III) or Ti (II). At the other end, the only oxidation state of zinc is \(+ 2\) (no d electrons are involved). The maximum oxidation states of reasonable stability correspond in value to the sum of the s - and d-electrons upto manganese ( \(Ti ^{I V} O _2, V^V O _2^{+}, Cr ^{V I} O _4^{2-}\), \(Mn ^{\text {VII }} O _4\) ) followed by a rather abrupt decrease in stability of higher oxidation states, so that the typical species to follow are \(Fe ^{I I, I I I}, Co ^{I I, I I I}, Ni ^{I I}, Cu ^{I, I I}, Zn ^{I I}\).
The variability of oxidation states, a characteristic of transition elements, arises out of incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity, e.g. \(V ^{I I}, V^{I I I}, V^{I V}, V^V\). This is in contrast with the variability of oxidation states of non transition elements where oxidation states normally differ by a unit of two.
An interesting feature in the variability of oxidation states of the d-block elements is noticed among the groups (groups 4 through 10). Although in the p-block, the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of dblock. For example, in group 6, \(Mo ( VI )\) and \(W ( VI )\) are found to be more stable than \(Cr ( VI )\). The \(Cr ( VI )\) in the form of dichromate in acidic medium is a strong oxidising agent, whereas \(MoO _3\) and \(WO _3\) are not.
Low oxidation states are found when a complex compound has ligands capable of \(\pi\)-acceptor character in addition to the \(\sigma\)-bonding. For example, in \(Ni ( CO )_4\) and \(Fe ( CO )_5\), the oxidation state of nickel and iron is zero.
Identify the stable pair of ions :

Arrange the following spermatogenic cells in correct sequence of spermatogenesis.
(A) Spermatids
(B) Spermatogonium
(C) Secondary Spermatocytes
(D) Primary Spermatocytes
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The integrating factor of the linear differential equation \(\frac{d y}{d x}+P(x) y=Q(x)\) is:

If a pink flowered Snapdragon plant is crossed with a white flowered Snapdragon plant, what will be the phenotypic ratio of their progeny?

Resolving power of a telescope can be increased by increasing:

Read the following paragraph and answer the given questions.
Cancer detection is based on biopsy and histopathological studies. Techniques like radiography (use of X-rays), CT (computed tomography) and MRI (magnetic resonance imaging) are very useful to detect cancers of the internal organs. Do you know why mother's milk is considered very essential for the new-born infant? The yellowish fluid colostrum secreted by mother during the initial days of lactation has abundant antibodies (IgA) to protect the infant. Different species of Plasmodium (P. vivax, P. malaria and P. falciparum) are responsible for different types of malaria. Of these, malignant malaria caused by Plasmodium falciparum is the most serious one and can even be fatal. The exaggerated response of the immune system to certain antigens present in the environment is called allergy. The use of drugs like anti-histamine, adrenalin and steroids quickly reduce the symptoms of allergy. Drugs like barbiturates, amphetamines, benzodiazepines, and other similar drugs, that are normally used as medicines to help patients cope with mental illnesses like depression and insomnia, are often abused. Morphine is a very effective sedative and painkiller, and is very useful in patients who have undergone surgery.
Which type of drug cannot quickly reduce the symptoms of allergy?

Which of the following statements are correct?
(A) If \(E\) and \(F\) are independent events then \(P(E \cap F)=P(E) P(F)\)
(B) If \(E\) and \(F\) are mutually exclusive events, then \(P(E \cup F)=P(E)+P(F)-P(E) P(F)\)
(C) The conditional probability of an event \(E\), given the occurrence of the event \(F\) is given by \(\frac{P(E \cap F)}{P\left(F^{\prime}\right)}, P(F) \neq 0\)
(D) If \(E\) and \(F\) be the events associated with the sample space \(S\) of an experiment, then \(P(\bar{E} \mid F)=2-P(E \mid F)\)
Choose the correct answer from the options given below: