NEET Practice Questions (MCQs) with Answers & Solutions

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Why is it not possible to form a p-n junction by simply joining a p-type semiconductor slab with an n-type semiconductor slab physically?

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Explanation

From Example 14.3, 'No! Any slab, howsoever flat, will have roughness much larger than the inter -atomic crystal spacing (~2 to 3 Ã…) and hence continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers.'

What constitutes the 'depletion layer' in a p-n junction?

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Explanation

Under 'POINTS TO PONDER', point 12 mentions, 'When such a junction is made, a ‘depletion layer’ is formed consisting of immobile ion-cores devoid of their electrons or holes. This is responsible for a junction potential barrier.'

What is the typical order of thickness of the depletion region in a p-n junction?

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Explanation

The NCERT text states, 'The thickness of depletion region is of the order of one-tenth of a micrometre.'

The primary cause for the flow of charge carriers in the drift current during p-n junction formation is:

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Explanation

The text explains, 'Due to this field, an electron on p-side of the junction moves to n-side and a hole on n-side of the junction moves to p-side. The motion of charge carriers due to the electric field is called drift.'

What is the polarity of the potential established across a p-n junction at equilibrium?

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Explanation

The NCERT states, 'The n-material has lost electrons, and p material has acquired electrons. The n material is thus positive relative to the p material.'

Which of the following elements exhibits the largest number of oxidation states in the d-block series?

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Explanation

The 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. (NCERT, Chapter: The d - and f - Block Elements, Page 97)

Why is scandium(II) virtually unknown?

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Explanation

The lesser number of oxidation states at the extreme ends stems from either too few electrons to lose or share (Sc, Ti). Thus, early in the series scandium(II) is virtually unknown. (NCERT, Chapter: The d - and f - Block Elements, Page 97)

Which of the following statements about zinc's oxidation state is correct?

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Explanation

At the other end (of the series), the only oxidation state of zinc is +2 (no d electrons are involved). (NCERT, Chapter: The d - and f - Block Elements, Page 97)

The maximum oxidation states of reasonable stability for transition elements correspond to the sum of s and d electrons up to which element?

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Explanation

The maximum oxidation states of reasonable stability correspond in value to the sum of the s and d electrons upto manganese ($Ti^{IV}O_2$, $V^{V}O_2^+$, $Cr^{VI}O_4^{2–}$, $Mn^{VII}O_4^{–}$). (NCERT, Chapter: The d - and f - Block Elements, Page 97)

The variability of oxidation states in transition elements arises from:

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Explanation

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., $V^{II}$, $V^{III}$, $V^{IV}$, $V^{V}$. (NCERT, Chapter: The d - and f - Block Elements, Page 97)

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