alleryStart learning →Which is the strongest oxidizer in halogens?
- Br₂
- F₂ (E° = +2.87 V, highest)
- I₂
- Cl₂
Show answer and solution
Correct answer: B. F₂ (E° = +2.87 V, highest)
**Standard reduction potentials:** F₂ > Cl₂ > Br₂ > I₂.
F₂ is strongest oxidizer in chemistry.
Indian Navy SSR (Sailor) JEE Main 2023 Mixed — practice questions
Explore 32 free JEE Main 2023 Mixed questions in Indian Navy SSR (Sailor) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.
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Showing questions 1–30 of 32.
Which is the strongest oxidizer in halogens?Read the answer and worked solution →For a hydrogen-like atom of nuclear charge $Z$, the radius of the $n$th Bohr orbit scales asRead the answer and worked solution →An electron is accelerated through a potential difference of $100\,V$. Its de Broglie wavelength is approximatelyRead the answer and worked solution →As per your syllabus, what is the MOLECULAR MASS of methane (CH₄) — using C = 12.011 u and H = 1.008 u?Read the answer and worked solution →As per your syllabus, the charge-to-mass ratio e/m_e of the electron, measured by J.J. Thomson in 1897, is which value?Read the answer and worked solution →The de Broglie wavelength of a particle with momentum p is:Read the answer and worked solution →Order the atomic models as they were proposedRead the answer and worked solution →Match each scientist to their modelRead the answer and worked solution →Match each model to what it could NOT explainRead the answer and worked solution →True or false: Rutherford's modelRead the answer and worked solution →True or false: Bohr's modelRead the answer and worked solution →Complete the reason Bohr's model failsRead the answer and worked solution →Which model does this belong to?Read the answer and worked solution →Match each quantum number to what it is calledRead the answer and worked solution →True or false: the quantum mechanical modelRead the answer and worked solution →Order the structure: from atom down to electronRead the answer and worked solution →Complete the (n + l) ruleRead the answer and worked solution →True or false: orbital energiesRead the answer and worked solution →Match each rule to what it forbids or requiresRead the answer and worked solution →Tap every statement Hund's rule requiresRead the answer and worked solution →The fundamental particlesRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Which principle is being described?Read the answer and worked solution →Flip and pair: symbol with meaningRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Tap every statement true of the Schrodinger modelRead the answer and worked solution →Reading the models backRead the answer and worked solution →Assertion and reason: filling orbitalsRead the answer and worked solution →
Worked practice questions
For a hydrogen-like atom of nuclear charge $Z$, the radius of the $n$th Bohr orbit scales as
- $r_n \propto nZ$
- $r_n \propto n^2/Z$
- $r_n \propto Z/n^2$
- $r_n \propto n/Z^2$
Show answer and solution
Correct answer: B. $r_n \propto n^2/Z$
$r_n = (0.529\,Å)\,n^2/Z$.
**So radius grows as $n^2$ but shrinks as $Z$ — explaining why He$^+$ has a smaller $n=1$ orbit than H.**
An electron is accelerated through a potential difference of $100\,V$. Its de Broglie wavelength is approximately
- $1.23\,Å$
- $12.3\,Å$
- $0.0123\,Å$
- $0.123\,Å$
Show answer and solution
Correct answer: A. $1.23\,Å$
For an electron accelerated through $V$ volts, $\lambda = \dfrac{12.27}{\sqrt{V}}\,Å$.
With $V = 100$: $\lambda = 12.27/10 \approx 1.23\,Å$.
This is comparable to atomic spacings — basis of electron diffraction.
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